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8.2: Addition and Multiplication Principles


Recall that the cardinality of a finite set (A), denoted (|A|), is the number of elements it contains.

Example (PageIndex{1}label{eg:addmult-01})

If (A={-1,0,2}), then (|A|=3). Also, [egin{array}{rcl} |{2}| &=& 1, |{2,5,-1,-3}| &=& 4, |{xinmathbb{R}mid x^2=1}| &=& 2. end{array} onumber] Notice that (|emptyset|=0), because an empty set does not contain any element.

It becomes more interesting when we consider the cardinality of a union or an intersection of two or more sets.

Example (PageIndex{2}label{eg:addmult-02})

Determine (|A cup B|) and (|A cap B|) if (A={2,5}) and (B={7,9,10}).

Solution

Since (Acup B = {2,5,7,9,10}), and (Acap B = emptyset), it is clear that (|Acup B|=5), and (|Acap B|=0).

Example (PageIndex{3}label{eg:addmult-03})

Determine (|A cup B|) and (|A cap B|) if (A={2,5}) and (B={5,9,10}).

Solution

Since (Acup B = {2,5,9,10}), and (Acap B = {5}), it is clear that (|Acup B|=4), and (|Acap B|=1).

hands-on exercise (PageIndex{1}label{he:addmult-01})

Let (A={ninmathbb{Z} mid -5leq nleq3}), and (B={ninmathbb{Z} mid -3leq nleq5}). Evaluate (|Acap B|) and (|Acup B|).

The difference between the last two examples is whether the two sets (A) and (B) have a nonempty intersection. Two sets (A) and (B) are disjoint if (A cap B = emptyset). A collection of sets (A_1, A_2, ldots, A_n) is said to be pairwise disjoint if (A_i cap A_j = emptyset) whenever (i eq j). When (A_1, A_2, ldots, A_n) are pairwise disjoint, their union is called a disjoint union.

Example (PageIndex{4}label{eg:addmult-04})

Let (A={1,0,-1}), (B={-2,0,2}), (C={-2,2}) and (D={3,4,5}). Then (A), (C), and (D) are pairwise disjoint, so are (B) and (D), but (A), (B), and (C) are not.

Theorem (PageIndex{1}): Addition Principle

If the finite sets (A_1, A_2, dots, A_n) are pairwise disjoint, then [|A_1 cup A_2 cup cdots cup A_n| = |A_1| + |A_2| + cdots + |A_n|. onumber]

Use the addition principle if we can break down the problems into cases, and count how many items or choices we have in each case. The idea is, instead of counting a large set, we divide it up into several smaller subsets, and count the size of each of them. The cardinality of the original set is the sum of the cardinalities of the smaller subsets. This divide-and-conquer approach works perfectly only when the sets are pairwise disjoint.

Example (PageIndex{5}label{eg:addmult-05})

To find the number of students present at a lecture, the teacher counts how many students there are in each row, then adds up the numbers to obtain the total count.

When the sets are not disjoint, the addition principle does not give us the right answer because the elements belonging to the intersection are counted more than once. We have to compensate the over-counting by subtracting the number of times these elements are over-counted. The simplest case covers two sets.

Theorem (PageIndex{2}label{pie}): Principle of Inclusion-Exclusion (PIE)

For any finite sets (A) and (B), we have [|Acup B| = |A|+|B|-|Acap B|. onumber]

Proof

Observe that (Acup B) is the disjoint union of three sets [A cup B = (A-B) cup (A cap B) cup (B-A). onumber] It is clear that (|A-B| = |A|-|Acap B|), and (|B-A|=|B|-|Acap B|). Therefore, [egin{array}{rcl} |A cup B| &=& |A-B| + |Acap B| + |B-A| &=& (|A|-|A cap B|) + |Acap B| + (|B|-|A cap B|) &=& |A| + |B| - |A cap B|, end{array} onumber] which is what we have to prove.

The principle of inclusion-exclusion also works if (A) and (B) are disjoint, because in such an event, (|Acap B|=0), reducing PIE to the addition principle.

Example (PageIndex{6}label{eg:pie})

Assume the current enrollment at a college is 4689, with 60 students taking MATH 210, 42 taking CSIT 260, and 24 taking both. Together, how many different students are taking these two courses? In other words, determine the number of students who are taking either MATH 210 or CSIT 260.

Solution

Let (A) be the set of students taking MATH 210, and (B) the set of students taking CSIT 260, Then, (|A|=60), (|B|=42), and (|Acap B|=24). We want to find (|Acup B|). According to PIE, [|Acup B| = |A| + |B| - |Acap B| = 60+42-24 = 78. onumber] Therefore, 78 students are taking either MATH 210 or CSIT 260.

Example (PageIndex{7}label{eg:addmult-07})

Among 4689 students, 2112 of them have earned at least 60 credit hours and 2678 of them have earned at most 60 credit hours. How many students are there who have accumulated exactly 60 hours?

Solution

Let (A) be the set of students who have earned at least 60 credit hours, and (B) be the set of students who have earned at most 60 credit hours. We want to find (|Acap B|). According to PIE, [4689 = |Acup B| = |A|+|B|-|Acap B| = 2112+2678-|Acap B|. onumber] Hence, [|Acap B| = (2112+2678)-4689 = 101. onumber] There are 101 students who have accumulated exactly 60 credit hours.

hands-on exercise (PageIndex{2}label{he:addmult-02})

The attendance at two consecutive college football games was 72397 and 69211 respectively. If 45713 people attended both games, how many different people have watched the games?

hands-on exercise (PageIndex{3}label{he:addmult-03})

The attendance at two consecutive college football games was 72397 and 69211 respectively. If 93478 different individuals attended these two games, how many have gone to both?

Sometimes, it is easy to work with the complement of a set.

Lemma (PageIndex{3})

For any finite set (S), we have [|overline{S}| = |{cal U}| - |S|, onumber] where ({cal U}) is the universal set containing (S).

Example (PageIndex{8}label{eg:addmult-08})

In Example 8.2.6, since there are 78 students taking either MATH 210 or CSIT 260, the number of students taking neither is (4689-78=4611).

The principle of inclusion-exclusion can be extended to any number of sets. The situation is more complicated, because some elements may be double-counted, some triple-counted, etc. To give you a taste of the general result, here is the principle of inclusion-exclusion for three sets.

Theorem (PageIndex{1})

For any three finite sets (A), (B) and (C), [|A cup B cup B| = |A| + |B| + |C| - |A cap B| - |A cap C| - |B cap C| + |A cap B cap C|. onumber]

Proof

The union (A cup B cup C) is the disjoint union of seven subsets: [displaylines{ A-(Bcup C), quad B-(Ccup A), quad C-(Acup B), quad (Acap B)-(Acap Bcap C), cr (Bcap C)-(Acap Bcap C), quad (Ccap A)-(Acap Bcap C), quad mbox{and}quad Acap Bcap C. cr} onumber] We can apply an argument similar to the one used in the union of two sets to complete the proof. We leave the details as an exercise.

hands-on exercise (PageIndex{4}label{he:addmult-04})

A group of students claims that each of them had seen at least one part of the Back to the Future trilogy. A quick show of hands reveals that

  • 47 had watched Part I;
  • 43 had watched Part II;
  • 32 had watched Part III;
  • 33 had watched both Parts I and II;
  • 27 had watched both Parts I and III;
  • 25 had watched both Parts II and III;
  • 22 had watched all three parts.

How many students are there in the group?

Another useful counting technique is the multiplication principle.

Theorem (PageIndex{5}) (Multiplication Principle)

For any finite sets (A) and (B), we have [|A imes B| = |A|cdot|B|. onumber]

Clearly, this can be extended to an (n)-fold Cartesian product.

Theorem (PageIndex{6})

For any finite sets (A_1, A_2, ldots, A_n), we have [|A_1 imes A_2 imes cdots imes A_n| = |A_1| cdot |A_2| cdot cdots cdot |A_n|. onumber]

In many applications, it may be helpful to use an equivalent form.

Theorem (PageIndex{7}) (Multiplication Principle: Alternate Form)

If a task consists of (k) steps, and if there are (n_i) ways to finish step (i), then the entire job can be completed in (n_1 n_2 ldots n_k) different ways.

Now that we have two counting techniques, the addition principle and the multiplication principle, which one should we use? The major difference between them is whether

  • the jobs can be divided into cases, groups, or categories; or
  • each job can be broken up into steps.

In practice, it helps to draw a picture of the configurations that we are counting.

Example (PageIndex{9}label{eg:addmult-09})

How many different license plates are there if a standard license plate consists of three letters followed by three digits?

Solution

We need to decide how many choices we have in each position. Draw a picture to show the configuration. Draw six lines to represent the six positions. Above each line, describe briefly the possible candidates for that position, and under each line, write the the number of choices.

[ egin{array}{ccccccc} ext{choices:} & ext{any letter} & ext{any letter} & ext{any letter} & ext{any digit} & ext{any digit} & ext{any digit} & - & - & - & - & - & - ext{# of choices:} & 26 & 26 & 26 & 10 & 10 & 10 end{array} onumber]

This left-to-right configuration suggests that the multiplication principle should be used. The answer is (26cdot 26cdot 26cdot 10cdot 10cdot 10 = 260^3).

As you become more experienced, you can argue directly, as follows. There are 26 choices for each of the three letters, and 10 choices for each digit. So there are (26cdot 26cdot 26cdot 10cdot 10cdot 10 = 260^3) different license plates.

Example (PageIndex{10}label{eg:addmult-10})

Find the number of positive integers not exceeding 999 that end with 7.

Solution 1

The integers can have one, two, or three digits, so we have to analyze three cases.

  • Case 1. There is only one integer with one digit, namely, the integer 7.
  • Case 2. If there are two digits, the first could be any digit between 1 and 9, and the last digit must be 7. [ egin{array}{ccc} ext{choices:} & 1-9 & 7 & - & - ext{# of choices:} & 9 & 1 end{array} onumber] This gives us nine choices.
  • Case 3. If there are three digits, the first digit could be any digit between 1 and 9, the second any digit between 0 and 9, and the last digit must be 7. [ egin{array}{cccc} ext{choices:} & 1-9 & ext{any digit} & 7 & - & - & - ext{# of choices:} & 9 & 10 & 1 end{array} onumber] Hence, there are 90 integers in this case.

Combining the three cases, we have a total of (1+9+90=100) integers that meet the requirements.

Solution 2

The integers could be written as three-digit integers if we allow 0 as the leading digits. For instance, 7 can be written as (007), and (34) as (034). Under this agreement, we have to fill three positions where the last one is always occupied by the digit 7. The first two digits are (0, 1, 2, ldots, 8), or 9, so there are 10 choices for each position.

[ egin{array}{cccc} ext{choices:} & ext{any digit} & ext{any digit} & 7 & - & - & - ext{# of choices:} & 10 & 10 & 1 end{array} onumber]

Together, there are (10cdot10=100) such integers.

hands-on exercise (PageIndex{5}label{he:addmult-05})

How many natural numbers less than 1000000 are there that end with the digit 3?

hands-on exercise (PageIndex{6}label{he:addmult-06})

How many natural numbers less than 10000 are there that end with the digit 0?

Example (PageIndex{11}label{eg:addmult-11})

Determine the number of four-digit positive integers without repeated digits.

Solution

We want to determine how many choices there are for each place value. The first digit has nine choices because it cannot be 0. Once the first digit is chosen, there are nine choices left for the second digit; and then eight choices for the next digit, and seven choices for the last digit. Together, we have (9cdot 9cdot 8cdot 7 =4536) four-digit positive integers that do not contain any repeated digits. Question: Can we start counting from the last digit?

hands-on exercise (PageIndex{7}label{he:addmult-07})

How many six-digit natural numbers are there that do not have any repeated digit?

Example (PageIndex{12}label{eg:addmult-12})

Determine (|wp(S)|), where (S) is an (n)-element set.

Solution

We want to determine the number of ways to form a subset. Let the (n) elements be (s_1, s_2, ldots, s_n). To form a subset, we go through each element (s_i) and decide whether it should be included in the subset, thus there are two choices for each element.

[ egin{array}{ccccc} ext{element:} & s_1 & s_2 & & s_n ext{choices} & ext{Y/N} & ext{Y/N} & cdots & ext{Y/N} & - & - & & - ext{# of choices:} & 2 & 2 & & 2 end{array} onumber]

We have (underbrace{2cdot2cdot,cdots, cdot2}_{n ext{ factors}} = 2^n) ways to form the subsets. Thus, (|wp(S)|=2^n).

Example (PageIndex{13}label{eg:addmult-13})

How many two-digit positive integers do not have consecutive 5s?

Solution 1

There are three disjoint cases:

  1. both digits are not 5,
  2. only the first digit is 5, and
  3. only the last digit is 5.

There are (8cdot9+9+8=89) integers that meet the requirement.

Solution 2

An easier solution is to consider the complement of the problem. There is only one integer with consecutive 5s, namely, the integer 55. There are 90 two-digit integers, hence (90-1=89) of them do not have consecutive 5s.

hands-on exercise (PageIndex{8}label{he:addmult-08})

How many three-digit natural numbers are there that do not have consecutive 4s?

Example (PageIndex{14}label{eg:addmult-14})

In how many ways can we draw a sequence of three cards from a standard deck of 52 cards?

Solution

This is a trick question! The answer depends on whether we can return a drawn card to the deck. With replacement, the answer is (52^3); without replacement, it is (52 cdot 51 cdot 50).

Example (PageIndex{15}label{eg:addmult-15})

A standard New York State license plate consists of three letters followed by four digits. Determine the number of standard New York State license plates with K as the first letter or 8 as the first digit.

Solution

The keyword “or” suggests that we are looking at a union, hence, we have to apply PIE. We need to analyze three possibilities:

  • There are (26^2cdot10^4) license plates with K as the first letter.
  • There are (26^3cdot10^3) license plates with 8 as the first digit.
  • There are (26^2cdot10^3) license plates with K as the first letter and 8 as the first digit.

The answer is (26^2cdot10^4+26^3cdot10^3-26^2cdot10^3).

hands-on exercise (PageIndex{9}label{he:addmult-09})

To access personal account information, a customer could log in to the bank’s web site with a PIN consisting of two letters followed by

  1. exactly four digits,
  2. at most six digits,
  3. at least two but at most 6 digits.

How many different PINs are there in each case?

Summary and Review

  • Use the addition principle if the problem can be divided into cases. Make sure the cases do not overlap.
  • If the cases overlap, the number of objects belonging to the overlapping cases must be subtracted from the total to obtain the correct count.
  • In particular, the principle of inclusion-exclusion states that (|Acup B|=|A|+|B|-|Acap B|).
  • Use the multiplication principle if the problem can be solved in several steps.
  • How can we get started? Imagine you want to list all the possibilities, what is a systematic way of doing so? Follow the steps, and count how many objects you would end up with.
  • It may be helpful to use a schematic diagram. Draw one line for each step. Above the lines, write the choices. Below the lines, write the number of choices. Apply the multiplication principle to finish the problem.
  • If there are other cases involved, repeat, and add the results from all the possible cases.

exercise (PageIndex{1}label{ex:addmult-01})

A professor surveyed the 98 students in her class to count how many of them had watched at least one of the three films in The Lord of the Rings trilogy. This is what she found:

  • 74 had watched Part I;
  • 57 had watched Part II;
  • 66 had watched Part III;
  • 52 had watched both Parts I and II;
  • 51 had watched both Parts I and III;
  • 45 had watched both Parts II and III;
  • 43 had watched all three parts.

How many students did not watch any one of these three movies?

exercise (PageIndex{2}label{ex:addmult-02})

Forty-six students in a film class told the professor that they had watched at least one of the three films in The Godfather trilogy. Further inquiry led to the following data:

  • 41 had watched Part I;
  • 37 had watched Part II;
  • 33 had watched Part III;
  • 33 had watched both Parts I and II;
  • 30 had watched both Parts I and III;
  • 29 had watched both Parts II and III.
  1. How many students had watched all three films?
  2. How many students had watched only Part I?
  3. How many students had watched only Part II?
  4. How many students had watched only Part III?

exercise (PageIndex{3}label{ex:addmult-03})

Joe has 10 dress shirts and seven bow ties. In how many ways can he match the shirts with bow ties?

exercise (PageIndex{4}label{ex:addmult-04})

A social security number is a sequence of nine digits. Determine the number of social security numbers that satisfy the following conditions:

  1. There are no restrictions.
  2. The digit 8 is never used.
  3. The sequence does not begin or end with 8.
  4. No digit is used more than once.

exercise (PageIndex{5}label{ex:addmult-05})

A professor has seven books on discrete mathematics, five on number theory, and four on abstract algebra. In how many ways can a student borrow two books not both on the same subject?

Hint

Which two subjects would the student choose?

exercise (PageIndex{6}label{ex:addmult-06})

How many different collections of cans can be formed from five identical Cola-Cola cans, four identical Seven-Up cans, and seven identical Mountain Dew cans?

Hint

How many cans of Cola-Cola, Seven-Up, and Mountain Dew would you pick?

exercise (PageIndex{7}label{ex:addmult-07})

How many five-letter words (technically, we should call them strings, because we do not care if they make sense) can be formed using the letters A, B, C, and D, with repetitions allowed. How many of them do not contain the substring BAD?

Hint

For the second question, consider using a complement.

exercise (PageIndex{8}label{ex:addmult-08})

How many different five-digit integers can be formed using the digits 1, 3, 3, 3, 5?

Hint

The three digits 3 are identical, so we cannot tell the difference between them. Consequently, what really matters is where we put the digits 1 and 5. Once we place the digits 1 and 5, the remaining three positions must be occupied by the digits 3.

exercise (PageIndex{9}label{ex:addmult-09})

Four cards are chosen at random from a standard deck of 52 playing cards, with replacement allowed. This means after choosing a card, the card is return to the deck, and the deck is reshuffled before another card is selected at random. Determine the number of such four-card sequences if

  1. There are no restrictions.
  2. None of the cards can be spades.
  3. All four cards are from the same suit.
  4. The first card is an ace and the second card is not a king.
  5. At least one of the four cards is an ace.

exercise (PageIndex{10}label{ex:addmult-10})

Three different mathematics final examinations and two different computer science final examinations are to be scheduled during a five-day period. Determine the number of ways to schedule these final examinations from 11 AM to 1 PM if

  1. There are no restrictions.
  2. No two examinations can be scheduled on the same day.
  3. No two examinations from the same department can be scheduled on the same day.
  4. Each mathematics examination must be the only examination for the day on which it is scheduled.

exercise (PageIndex{11}label{ex:addmult-11})

Determine the number of four-digit positive integers that satisfy the following conditions:

  1. There are no restrictions.
  2. No integer contains the digit 8.
  3. Every integer contains the digit 8 at least once.
  4. Every integer is a palindrome (A positive integer is a palindrome if it remains the same when read backward, for example, 3773 and 47874).

exercise (PageIndex{12}label{ex:addmult-12})

A box contains 12 distinct colored balls (for instance, we could label them as 1, 2, …, 12 to distinguish them). Three of them are red, four are yellow, and five are green. Three balls are selected at random from the box, with replacement. Determine the number of sequences that satisfy the following conditions:

  1. There are no restrictions.
  2. The first ball is red, the second is yellow, and the third is green.
  3. The first ball is red, and the second and third balls are green.
  4. Exactly two balls are yellow.
  5. All three balls are green.
  6. All three balls are the same color.
  7. At least one of the three balls is red.

exercise (PageIndex{13}label{ex:addmult-13})

Let (A={a,b,c,d,e,f}) and (B={1,2,3,4,5,6,7,8}). Determine the number of functions (f: A o B) that satisfy the following conditions:

  1. There are no restrictions.
  2. (f) is one-to-one.
  3. (f) is onto.
  4. (f(x)) is odd for at least one (x) in (A).
  5. (f(a)=3) or (f(b)) is odd.
  6. (f^{-1}(4)={a}).

exercise (PageIndex{14}label{ex:addmult-14})

How many onto functions are there from an (n)-element set (A) to ({a,b})?


Probability Addition and Multiplication Principles of Counting - PowerPoint PPT Presentation

Standard: MM1D1a - a. Apply the addition and multiplication principles of counting. Probability Addition and Multiplication Principles of Counting &ndash PowerPoint PPT presentation

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2 Answers 2

What is the number of ways of choosing four cards of the same suit when four cards are selected from a standard deck of $52$ playing cards?

Method 1: To be successful, we must choose four of the thirteen clubs or four of the thirteen diamonds or four of the thirteen hearts or four of the thirteen spades, which can be done in $inom<13> <4>+ inom<13> <4>+ inom<13> <4>+ inom<13> <4>= 4inom<13><4>$ ways.

The Addition Principle states that if there are $m$ ways of performing one task and $n$ ways of performing another task that cannot be done at the same time, then the number of ways of performing one of the tasks is $m + n$.

In this case, we may choose four clubs or four diamonds or four hearts or four spades. Since these tasks are mutually exclusive (cannot be performed at the same time), we add the number of ways each task can be performed.

The word or is an indication that you need to add.

Method 2: To be successful, we must choose one of the four suits and four of the thirteen cards of that suit in $inom<4><1>inom<13><4>$ ways.

The Multiplication Principle states that if one task can be performed in $m$ ways and a second task can be performed independently of the first in $n$ ways, then there are $mn$ ways of performing both tasks.

In this case, we choose a suit and then choose four cards of that suit. Since the number of ways we can select four cards of the chosen suit is independent of the choice of suit, we multiply the number of ways each task can be performed.

The word and is an indication that you have to multiply.

What is the number of ways of choosing four cards of different suits when four cards are selected from a standard deck of $52$ cards?

We must choose one of the thirteen clubs and one of the thirteen diamonds and one of the thirteen hearts and one of the thirteen spades, which can be done in $inom<13><1>inom<13><1>inom<13><1>inom<13><1>$ ways.

Notice that each of the four tasks is performed independently of the others, so we multiply the number of ways each task can be performed.

What is the number of ways of choosing two red and two black cards when four cards are selected from a standard deck of $52$ playing cards?

We must select two of the twenty-six red cards and two of the twenty-six black cards, which can be done in $inom<26><2>inom<26><2>$ ways.

Notice that each task is performed independently of the other, so we multiply the number of ways each task can be performed.

You first selected $26$ of the $52$ cards in the deck, then multiplied by the number of ways to select two of those $26$ cards of one of the colors. However, $inom<52><26>$ is the number of ways of selecting any $26$ cards in the deck (which do not all need to be the same color). Instead, you should have chosen all $26$ cards of that color and then chosen two cards of that color for each color, which can be done in $inom<26><26>inom<26> <2>= inom<26><2>$ ways.

Moreover, you have to this for the red cards and the black cards, so you have to multiply the number of ways of doing this for each color.


8.2: Addition and Multiplication Principles

If a_n is the number of widgets of size n and A(q) = a_0 + a_1 q + a_2 q^2 + a_3 q^3 + . then we call A(q) a generating function for widgets (of size n).
Similarly if b_n = the number of doodles of size n and B(q) = b_0 + b_1 q + b_2 q^2 + . is a generating function for doodles (of size n), then we have the following results:

The Additional Principle of generating functions:
A(q) + B(q) is the a generating function for the disjoint union of widgets and doodles (of size n).


The Mulitiplication Principle of generating functions:
A(q) B(q) is a generating function for pairs whose first element is a widget and second element is a doodle and the size of a pair is the sum of the size of the widget plus the size of the doodle.



This can be taken further. We can also look at other expressions of generating functions and say what they are in general:

A(q)/(1-q) is a generating function for the widgets of size less than or equal to n.

1/(1-qA(q)) is a generating function for 'sequences of widgets' (or k-tuples of widgets) where the size of a sequence of a widget is the number of widgets in the sequence (or k) plus the sum of the sizes of each of the widgets in the sequence.


Math Challenge

MDAS Rule is actually a rule to follow when we are going to solve a series of operations, that is the four fundamental operations of real numbers. MDAS rule stands for MULTIPLICATION, DIVISION, ADDITION and SUBTRACTION.

How are we going to use this rule?

First we have to perform all multiplications and divisions before performing the additions and subtraction.
In performing all multiplications and divisions, we should solve or read it from left to right. Do not try to solve any multiplication or division since it is easier or convenient to you to solve. Always solve from left to right whatever that comes first.
Follow the same process for additions and subtraction though we can apply the rule of signed numbers or we can also use the commutative law of addition.

1) 10 x 5 / 25 + 8 - 2 x 2 + 4 - 3
In solving this we perform first 10 x 5 together with 2 x 2. Giving us
50/25 + 8 - 4 + 4 - 3
Then we have to solve 50/25 before performing the addition and subtraction. So we have,
2 + 8 - 4 + 4 - 3 which is equal to
10 - 4 + 4 - 3. And the final answer is
7.

Try these:


1) 15 + 35/5 x 8 + 12 - 7 x 4 / 2 - 6
2) 105 / 3 - 35 x 2 + 35 - 70/2 + 35
3) 50 x 3 + 50/2 - 25 + 25/5 x 10

129 (na) komento:

I thougth after multiplication and division, it is ADDITION first then SUBTRACTION? Then it should have been:

you shouldn't add -4+4 as 8 it should be 0

parang (minus) 4 + 4 yun, hindi negative 4 + 4

no, at ease lang ang sign. tsaka wala namang parenthesis.

MDAS as I've understood it today: in solving such problem, MDAS indeed applies yet.

when it comes to addition and subtraction, addition is not really necessary should be conducted first then. Instead, if the signs left were minus and plus, simplify the factors into its sequence from left to right.

-4+4=8?
KAHIT CGURO GRADE 1 ALAM YAN EH

10 - 4 + 4 - 3 eto ba sinasabe moh ? wala nman parenthesis saka sabi sa article is from left to right.. d mo ba gets.

TAMA UNG SAGOT WAG KAYO MAG INASO

always apply the rules .. i am not good in mathematics but then i am aware of it ..

bear on mind always that there is a multiple intelligence possess by every individual ..

MDAS pwede ring DMSA. depende sa posisyon nga mga numbers. Mas madali lang talagang tandaan ang MDAS kaya yan ang laging ginagamit and then may kasunod na explanation:

How are we going to use this rule?

First we have to perform all multiplications and divisions before performing the additions and subtraction.
In performing all multiplications and divisions, we should solve or read it from left to right. Do not try to solve any multiplication or division since it is easier or convenient to you to solve. Always solve from left to right whatever that comes first.
Follow the same process for additions and subtraction though we can apply the rule of signed numbers or we can also use the commutative law of addition.

wew hahaha alam nyu ang M at D same lvl yan tsaka A at S kaya pwedi yan gani to ohhh DMSA wag kung san kayo mas gusto sumagot ginawa lang yang MDAS kasi madaling makabisado at mabasa

hnd ibg sbhn ng MDAS eh sunud sunod na. mas okay cguro na MD/AS.
Multiplication and division always comes first. followed by addition and subtraction. but if MD or AS is left on the operation, it should be left to right

1+1+2x3= alin po una? multiplication po ba?

Integers are not applied in that rule

A vendor buys pineapple at 11.50. He sells it for 15.00. How much was his gain if he was able to sell 25

ganito po yung comp nya.
10x5/25+8-2
x2+4-3=7
sol.
unahin ang multiplication
50/25+8-4+4-3
then division
2+8-4+4-3
since may rules sa addition and subtraction magiging ganito po sya.
addition
10-4+4-3 i solve lng po yung nasa left side
6+4-3
10-3
=7 comment kung mali po yung akin tnx.

CAS aka LaniXs IT-IIA.. hehe!

thanks for this informative post mam..

JBB AKA. JEFFREY B. BAQUERO.
I WILL OBEY STI MAASIN RULES.

I LIKE THE WAY YOU TEACH US MAAM..THANK YOU

PRINCESS STEFANI BAQUERO- IT 11A.
IM YOUR LADY MONSTER.

ung Tanong sa number 1 ay impossible
kasi tignan nyo
15 + 35/5 x 8 + 12 - 7 x 4 / 2 - 6
15 + 35 / 40 + 12 - 28 / 2 - 6
15 + + 12 - 14 - 6
pano nyo ddivide ung 35 / 40 ?
hahaha LMAO
xDD

you can interchange multiplication with division. so it doesn't really matter kung alin ang unahin mo sa kanilang dalawa. is it clear already?

unahin din kasi yong division. wag kang tumalon sa multiplication. peace.
elrica

"In performing all multiplications and divisions, we should solve or read it from left to right. Do not try to solve any multiplication or division since it is easier or convenient to you to solve. Always solve from left to right whatever that comes first."

15 + 35/5 x 8 + 12 - 7 x 4 / 2 - 6
15 + 7 x 8 + 12 - 28 / 2 - 6
15 + 56 + 12 - 14 - 6
63

in MDAS rule should the rule be followed accordingly from Multiplication to Division, Addition & Subtraction? if yes than i think problem 1 doesn't apply the rule. If No, than MDAS rule can be also DMAS rule to enable to solve problem 1. Correct me if i'm wrong. Thanks.

its true, that using MDAS multiplication should comes first before division. but that is not always the case. if multiplication and division are adjacent operations, the operation must be performed from left to right. meaning, the operation which comes first must be performed. same as addition and subtraction.

15 + 35/5 x 8 + 12 - 7 x 4 / 2 - 6
15 + 7 x 8 + 12 - 28 / 2 - 6
15 + 56 + 12 - 14 - 6
63

In performing all multiplications and divisions, we should solve or read it from left to right. Do not try to solve any multiplication or division since it is easier or convenient to you to solve. Always solve from left to right whatever that comes first.

always solve from left to right in solving problems involving multiplication and divisions? ang bobo mo naman . we can always start solving which is easier first especially if you know that it will save u time and effort . it will always give u the same answer.

ANG ANG MAUUNA PO KASI ISOLVE EH PAG MUMULTIPLY PO O DI KAYA PAG DIVIDE DEPEDENDE PO KUNG ANU NAUNA LEFT TO RYT GANUN DIN PO SA PAG ADD AND SUBTRACT KUNG ANU NAUNG OPERATION LEFT TO RYT

UNAHIN PO NATIN UNG PAG MULTIPLY O PAG DIVIDE DEPENDE SA NAUNA SA ORDER NA LEFT TO RYT

15 + 35 / 5 X 8 + 12 - 7 X 4 / 2 - 6
15 + 7 X 8 + 12 - 28 / 2 -6
15 + 56 + 12 - 14 - 6

NGAYON PO ADD AND SUBTRACT DEPENDE SA NAUNA SA LEF TO RYT ORDER

15 + 56 + 12 - 14 - 6
71 + 12 - 14 - 6
83 - 14 - 6
69 - 6
63

na totrololol tlqa utak sa MDAS . >.<

na totrololol tlqa utak ko sa MDAS . >.<

panu pag addition left to right parin ba?
kasi di ba pag addition right to left ang pag add,tama po ba?
sa MDAS alin po ang major operation sa kanila..bakit kailanga multiplication ung nauuna?

hahahaha klaro naman usapan sa MDAS laverne..
hiniwalay ang (multiplication at division)sa (additon at subtraction). so depende lang kong sino nauna(multiplication or division vise versa) basta left to right ang pag solve. ganon din sa addition at subtraction. in every rule, there is an exemption so ito na yon sa MDAS 35/5 X 8 = 7 X 8 = 56 kasi nauna ang division kaysa multiplication. clear?

wla poh bang matinung example kailangan koh poh kcing ma tutunan tung mdas kc poh. bobo aqou sa math ehh pro kung tutulungan nyo poh ako baka sakaling mag pursige poh ako upang ma tutu.

nais qou lng nman poh na mag bigay kau ng paraan kung pano ang teknect sa pag so solve. thnk you poh..

pag verbal exam po ba nasusunod pa ang mdas rule

pag verbal exam po ba nasusunod pa ang mdas rule

PROBLEM 1:
15 + 35 / 5 x 8 + 12 – 7 x 4 / 2 – 6 = ?
15 + (35 / 5) x 8 + 12 – (7 x 4) / 2 – 6 = ?
15 + 7 x 8 + 12 – 28 / 2 – 6 = ?
15 + (7 x 8) + 12 – (28 / 2) – 6 = ?
15 + 56 + 12 – 14 – 6 = ?
(15 + 56 + 12) – 14 – 6 = ?
83 – 14 – 6 = 63
ANS. 63

PROBLEM 2:
105 / 3 – 35 x 2 + 35 – 70 / 2 + 35 = ?
(105 / 3) – (35 x 2) + 35 – (70 / 2) + 35 = ?
35 – 70 + 35 – 35 + 35 = ?
(35 – 70) + 35 – 35 + 35 = ?
- 35 + 35 – 35 + 35 = ?
(- 35 + 35) – 35 + 35 = ?
0 – 35 + 35 = ?
- 35 + 35 = 0
ANS. 0

PROBLEM 3:
50 x 3 + 50 / 2 – 25 + 25 / 5 x 10 = ?
(50 x 3) + (50 / 2) – 25 + (25 / 5) x 10 = ?
150 + 25 – 25 + 5 x 10 = ?
150 + 25 – 25 + (5 x 10) = ?
150 + 25 – 25 + 50 = ?
(150 + 25) – 25 + 50 = ?
175 – 25 + 50 = ?
(175 – 25) + 50 = ?
150 + 50 = 200
ANS. 200

1) 15 + 35/5 x 8 + 12 - 7 x 4 / 2 - 6
= 15 + 7 x 8 + 12 - 28 / 2 - 6
= 15 + 56 + 12 - 14 - 6
= 71 + 12 - 14 - 6
= 83 - 14 - 6
= 69 - 6
= 63
or
= 15 + 7 x 8 + 12 - 28 / 2 - 6
= 15 + 56 + 12 - 14 - 6
= (15 + 56 + 12 ) - (14 + 6)
= 83 - 20
= 63
2) 105 / 3 - 35 x 2 + 35 - 70/2 + 35
= 35 - 70 + 35 - 35 + 35
= -35 + 35 - 35 + 35
= 0 - 35 + 35
= -35 + 35
= 0
or
= 35 - 70 + 35 - 35 + 35
= (35 + 35 + 35 ) - (70 + 35)
= 105 - 105
= 0
3) 50 x 3 + 50/2 - 25 + 25/5 x 10
= 150 + 25 - 25 + 5 x 10
= 150 + 25 - 25 + 50
= 175 - 25 + 50
= 150 + 50
= 200
or
= 150 + 25 - 25 + 5 x 10
= 150 + 25 - 25 + 50
= (150 + 25 + 50) - 25
= 225 - 25
= 200

what if kung addition and subtraction lang yung nakalagay parang ganto

That will be computed in this way:
2+4+7-5-4
=13-9
=4

if addition and subtraction are adjacent operations, the operation must be performed from left to right. meaning, the operation which comes first must be performed. same as multiplication and division

you should combine the similar terms then proceed to the operation ..

1) 10 x 5 / 25 + 8 - 2 x 2 + 4 - 3
In solving this we perform first 10 x 5 together with 2 x 2. Giving us
50/25 + 8 - 4 + 4 - 3
Then we have to solve 50/25 before performing the addition and subtraction. So we have,
2 + 8 - 4 + 4 - 3 which is equal to
10 - 4 + 4 - 3. And the final answer is
7.

correct if im wrong.
10 x 5 / 25 + 8 - 2 x 2 + 4 - 3
50 / 25 + 8 - 4 + 4 -3
2 + 8 - 4 + 4 - 3
10 - 8 -3
2 - 3
= -1

following the MDAS rule the answer is -1

that is: do the addition first, then use the negative (minus) sign and proceed to the next step. that should be written as - (4+4) which becomes -8. always consider the whole equation in performing the MDAS rule.

when you enclose the operation in parenthesis, the sign should be included inside. just like this:

2 + 8 - 4 + 4 - 3
(+2) + (+8) + (-4) + (+4) + (-3)
2 + 8 + (-4+4) + (-3)

Consider the expression Y subtracted by Z. We can write Y-Z as Y+(-Z)

correct if im wrong.
10 x 5 / 25 + 8 - 2 x 2 + 4 - 3
50 / 25 + 8 - 4 + 4 -3
2 + 8 - 4 + 4 - 3
10 - 8 -3
2 - 3
= -1

you're wrong
MDAS rule perform addition first before subtraction
so to be easy write this way
2+8+4-4-3=7
CLEAR?

kahit gamitan ng + and - deduction o kahit illustrate through drawing or kahit i rumble ang numbers without changing their signs. 7 pa rin ang sagot. sample:
combine + and - sign is = to 0
2=++
8=++++++++
4=++++
-4=----
-3=---
all + are 14 and all - are 7 +14-7= 7
kahit i drawing nyo pa :)

Para masigurado ang sagot, i-copy paste sa microsoft excel, maunang isulat ang = sign saka i-paste yung equation. palitan din ng * yunng x, saka i-enter. Sinusunod niyan ang MDAS rule.

ka ki kriya pahle karo ta piche do bhag guna karo dhan jor do rin ko do ghatay

HINDI MATHS
Raman Prakash Mishra
Patna

15 +35x8/5 (which is: 280/5) +12 -7x4/2 (which is: 28/2) -6

1) 10 x 5 / 25 + 8 - 2 x 2 + 4 - 3
In solving this we perform first 10 x 5 together with 2 x 2. Giving us
50/25 + 8 - 4 + 4 - 3
Then we have to solve 50/25 before performing the addition and subtraction. So we have,
2 + 8 - 4 + 4 - 3 which is equal to
10 - 4 + 4 - 3. And the final answer is
7.
NOTE: if we are going to follow the MDAS rule, the answer is -1.

MDAS: you should always consider the signs used in the equation. and remember the rules in performing the operations with LIKE SIGNS (i.e: +positive, +positive or -negative, -negative) - the answer is always +positive. with UNLIKE SIGNS (i.e: +positive, negative or vice versa) - the answer should always be in a negative sign.

=(10x5)/25+8- (2x2) +4-3 (here, do the MULTIPLICATION first, then proceed to the next step
=50/25+8-4+4-3 -DIVISION
=2+8-4+4-3 -ADDITION
=(2+8)- (4+4) -3 -Group together those with the same signs using a parenthesis and take note of the sign outside. then the last step -SUBTRACTION
=10-(8)-3 (thus, it becomes)
=10-8(unlike sign)-3
=2-3 (in determining what sign to be used, remember the rules in adding unlike signs. you should always use the sign of the larger number in determining the sign to be used in your final answer.
= -1

10x5/25+8-2x2+4-3
i may group it by parenthesis

[10x5]/25 + [8] - [2x2] + [4] - [3]
50/25 + [8] - [ 4 ] + [4] - [3]

bear in mind that inside the parenthesis are positive digits
[4] or [+4] // [3] or [+3], if outside the parenthesis is negative sign, the entire sign inside the parenthesis will change as well.
so it goes like this :


Choosing the Committee with 2 Women and 2 Men

Choosing the Committee with 3 Women and 1 Man

Choosing the Committee with 4 Women and 0 Men

EA1, EA2 and EA3 are Mutually Exclusive Events.

Occurrence of one of these events prevents the occurrence of the others. If the committee is chosen in one of these ways we can say that it was not chosen in the other ways.

Choosing 2 men from the total 5

Choosing 2 women from the total 6

By the fundamental counting theorem of multiplication

Number of ways in which the committee can be chosen with 2 women and 2 men

= (Number of ways in which the 1 st sub event of choosing the 2 men from a total 5 can be accomplished) × (Number of ways in which the 2 nd sub event of choosing the 2 women from a total 6 can be accomplished)

Choosing 1 man from the total 5

Choosing 3 women from the total 6

By the fundamental counting theorem of multiplication

Number of ways in which the committee can be chosen with 3 women and 1 man

= (Number of ways in which the 1 st sub event of choosing the 1 man from a total 5 can be accomplished) × (Number of ways in which the 2 nd sub event of choosing the 3 women from a total 6 can be accomplished)

Choosing 0 men from the total 5

Choosing 4 women from the total 6

By the fundamental counting theorem of multiplication

Number of ways in which the committee can be chosen with 4 women and 0 men

= (Number of ways in which the 1 st sub event of choosing 0 men from a total 5 can be accomplished) × (Number of ways in which the 2 nd sub event of choosing the 4 women from a total 6 can be accomplished)

By the fundamental counting theorem of addition,

The number of ways in which the committee of 4 members be chosen such that it consists of at least 2 women

= (Number of ways in which the 1 st alternative event of choosing the 2 men and 2 women can be accomplished) + (Number of ways in which the 2 nd alternative event of choosing the 1 man and 3 women can be accomplished) + (Number of ways in which the 3 rd alternative event of choosing the 0 men and 4 women can be accomplished)


8.2: Addition and Multiplication Principles

General Multiplication & Division:

  • Multiplication is Commutative (Margaret Carr)
  • Multiplication (Arrays) (LT)
  • Multiplication and Division Rules (Nicola Edwards)
  • T1 U9 Multiplication (David Arthur)
  • Division - Repeated Subtraction (Andrew Woodcock)
  • Division Snap (Nadine Turner) PDF
  • Gingerbread Multiplication (Sheena Florey) PDF
  • Toes Multiplication (Sheena Florey) PDF
  • Divisor Spiders Sheet 1 (Larissa Hughes) PDF Sheet 2 PDF
  • Arrays (Hamish Hobkinson) PDF
  • Function Machines (Mark Stallwood) PDF
  • Division (Amy Hedges) Bottom PDF - Middle PDF - Top PDF
  • Division Problems (Elaine Smith) DOC
  • Dividing Money by 4 & 5 (Joanne Gordon) DOC
  • Division Dominoes Game (Ashley Staniforth) Rules DOC - Dominoes DOC - Small DOC
  • Mental Methods Revision (Louise Pickering) DOC
  • Sums Square Multiplication Bingo (Campbell Airlie)
  • Inverse and Division (C. Williams) DOC
  • Division Dominoes (Joanne Nalton) DOC
  • Division Machines (Vicky Dowding) PDF
  • Choose the Product Games (Vicki Foy) PDF
  • Commutative Multiplication (Peter Smith) DOC
  • Division (by 2, 3, 5, 10) (Mark Wilson) DOC
  • Straight Division (Michelle Culliford) PDF
  • Straight Division 2 (Michelle Culliford) PDF
  • Using Known Facts ( BlkA U2) (Brenda Vaughan) DOC
  • Multiplication and Division Investigation (Kate Lowndes) DOC
  • Exploring Patterns in Linked Division Calculations (Louise Forster)
  • Exploring Patterns in Linked Division Calculations (Louise Forster) DOC
  • Multiplication Grids Starter (Paula Alty)
  • Division Grids Starter (Paula Alty)
  • Division Grids with Remainders Starter (Paula Alty)
  • Cooking Multiplication (Robert Bentall) DOC
  • Division Picture (Victoria Adams) DOC
  • Division Triangle Jigsaw (Peter Barnett) PDF
  • Division Hexagon Jigsaw (Peter Barnett) PDF
  • Missing Numbers: Multiplication & Division (Dhipa Begum) DOC
  • Multiplication Word problems (x2, x5, x10) (Cindy Shanks) DOC
  • Division Duck (4x) (Sarah Dickens)
  • Mental Multiplication Using Factors (Gemma Finbow)
  • Quick Multiplication Question Generator (R. Lovelock)
  • Multiplication "Jeopardy" Game (Helen Newton)
  • Multiplication Problems (Sheila Black) PDF
  • Multiplying two digits (Sheila Black) PDF
  • Multiplication (Ian Mason) Sheet 1 PDF - Sheet 2 PDF
  • Multiplication (Ian Mason) Sheet 3 PDF - Sheet 4 PDF
  • Factor Spiders (Larissa Hughes) Sheet 1 PDF Sheet 2 PDF
  • Division (Kevin Kerr) PDF
  • Division Spiders (Gareth Rein) DOC
  • Bear Multiplication (Judith Brayshaw) DOC
  • Multiplying and Dividing by 4 (Jon Fordham) DOC
  • Multiplication & Division Term 1 Unit 2 (Fred Daynes) Day 1 Day 2 Day 4 Day 5
  • Multiplication Worksheet Generator (Campbell Airlie)
  • Simple Multiplication Problems (C. Williams) DOC
  • Multiplication (Lots of) (Liz Hazelden) DOC
  • Multiplication & Division Relationship (Rich Robinson)
  • Multiplying by 4, 5 and 20 (Richard Queripel) DOC
  • Multiplication Games (Vicki Foy) PDF
  • Multiplication Questions (Anne Richard) DOC
  • Multiplication & Division Methods Poster (Ali McNamara) DOC
  • Y3 Division Test (Sharon Richard) DOC
  • Division Splat (Jim Usher) DOC
  • Associative Multiplication (Andrew Woodcock) XLS
  • Division Sheets (Linda Cook) DOC
  • Multiplying Measures (Michelle Culliford) PDF
  • Trio Triangles (Caroline Stares) DOC
  • Dividing 2 digit by 1 digit Numbers Mentally (Louise Forster) DOC
  • Pick and Match Calculations (Known Facts) (Louise Forster) DOC
  • Using Known Facts (Louise Forster) DOC
  • Animal Multiplication (Mez Miles) DOC
  • Octopus Multiples (Mez Miles) DOC
  • Using Multiplication Facts to Solve Division Questions (Paula Alty) DOC
  • Missing Numbers Division (Joanne Pooley)
  • Multiplication & Division Arrays (Emma Bagley) DOC
  • Linked Division (Louise Forster)
  • Multiple Monsters (Joanna Cayley)

Repeated Addition / Repeated Subtraction:

  • Repeated Addition as Multiplication (Katherine Gronert)
  • Repeated Addition & Multiplication (Raj Barard)
  • Multiplication as Repeated Addition (Amy Sheppard)
  • Multiplication Arrays (Julie Stead)
  • Division Using a Numberline (Toni Boucher)
  • Repeated Addition (Valerie Ryan)
  • Multiplication Build Up (Gill O'Neil) PDF
  • Changing Adding to Multiplication (Carol Wright) DOC
  • Division by Repeated Subtraction (Richard Queripel) DOC
  • Counter Array (Andy Cork) DOC
  • Groups of (Early Multiplication) (Shirley Lehmann)
  • Repeated Addition (Lesley Bratton)
  • Number Line for Repeated Addition (Morag Watson)
  • Division as Repeated Subtraction using a Number Line (Paula Alty) DOC
  • Multiplication & Division Arrays (Robert Jinkerson)
  • Division Using a Numberline (Kate Major)
  • Division by Repeated Subtraction (Chris Kirwan)
  • Multiplication (Repeated Addition and Arrays) (Andrea Harrison)
  • Division by Sharing (Avani Chotski)
  • Introduction to Multiplication (Jude Kuscher)
  • Introduction to Multiplication (Dawn Atkin) DOC
  • Division using repeated subtraction (Chez Owen) DOC
  • Patterns of Three (Tamsin Hall) PDF
  • Multiplying Sets (Carol Wright) DOC
  • Multiplication & Division on a Numberline (Ali McNamara) DOC
  • Dividing on a Numberline (Robert Jinkerson)
  • Division by Repeated Subtraction (with a numberline) (Richard Queripel) DOC
  • Multiplication Arrays (M A Crook) DOC
  • Counting in Groups (Lucy Hall) DOC

  • Division by Sharing (Claire Robinson)
  • Dividing and Sharing (3 levels) (Naomi Hass) DOC
  • Division by 3 (Rachael Durneen)
  • Sharing Counters (Liz Hazelden) DOC
  • Sharing Equally (Shirley Lehmann)
  • Smarties Share (Andy Cork) DOC
  • Simple Division Questions (Helen Bell) DOC
  • Division 'Rounding Remainders' (Michelle Rundle)
  • Sharing Sweets (1) - (2) - (3)
  • Sharing Zoo (Lesley Bratton)
  • Sharing Zoo (Lesley Bratton) DOC
  • Making Groups (Paul Smith)
  • Division as Sharing 1 (Carly Pitman)
  • Division as Sharing 2 (Carly Pitman)

  • Dividing Using Groups (Tim Pool)
  • Dividing by Grouping (Zoe Mayston) DOC
  • Division by Grouping with Remainders (Matt Lovegrove)
  • Division by Grouping (Vicky Frampton) DOC

Division with Remainders:

  • Division Word Problems with Remainders (Shazia Hussain) DOC
  • Division Picture PDF
  • Remainders as Fractions and Decimals (Richard Queripel) DOC
  • Rounding Up and Down After Division (David Peynado) DOC
  • Simple Divisions with Remainders (R. Lovelock)
  • Weetabix Division with Remainders (Heidi Morris-Duffin) DOC
  • Division with Remainders (Jenny Synnott)
  • Multiplication & Division (T3 Unit 2) (Joanne Robson) Day 1 Day 2
  • Simple Division with Remainders (C. Williams) DOC
  • Rounding Up After Division (Fiona Bell) DOC
  • Remainders Game (J. Balmer) DOC
  • Division Problems with Rounding (Helen Langford) DOC
  • Division with Remainders (Victoria Adams) DOC
  • Word Problems with Remainders (Steve Abey)

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Disciplemaking: Addition vs. Multiplication

Walter Henrichsen in Disciples Are Made Not Born suggests a hypothetical situation that clearly illustrates the process of multiplication. Suppose a father offers his two sons the choice of taking either one dollar a week for 52 weeks or one cent the first week and an amount each week for the next 51 one week’s that is double the previous week’s amount.

Which one would you choose?

The first choice would just be adding one dollar each week – that’s linear growth. At the end of 52 weeks – he’d have $52.

The second choice is multiplication – that’s exponential growth. If one of the sons chooses this, at the end of the year he will have an unbelievable amount of money. In fact, his allowance in the last week (not the total amount accumulated over 52 weeks) would be $22,517,998,136,852.48. Initially the multiplication is slow, but don’t let that deceive you. In the long run, addition never keeps pace with multiplication. Multiplication is explosive.

Another example: Suppose you start with a checkerboard of 64 squares. On the first square you place one grain of wheat. On the second square you place two grains, and on the third square you place four grains. How much wheat would you have to place on the last square if you continue doubling each succeeding square?

It would take enough wheat to cover India to a depth of fifty feet. The multiplication process is indeed explosive.

A disciplemaking ministry is built on exponential growth – the principle of multiplication. Consider a comparison of disciplemaking to mass evangelism:

Suppose you are a really great evangelist and you lead 1,000 people to faith in Christ every day. At the end of the first year there would be 365,000 new believers! That’s the principle of addition – 1,000 added every day.

Suppose another person in one year led one person to Christ and spent that year building his faith, teaching God’s truth, and training that individual to grow to maturity and to spiritually reproduce themselves in someone else. At the end of the first year there would be 2 disciplemakers.

If you continue to lead 1,000 people everyday to faith in Christ, at the end of the second year you’d have led 730,000 people to follow Jesus. The disciplemakers at the end of the second year will have invested themselves in two more individuals so that at the end of the year they are able to spiritually reproduce themselves. So after two years – there would be 4 disciplemakers.

If this process were to continue and you kept leading 1,000 people to Christ every day (adding 365,000 believers every year), at the end of 10 years you would have reached 3,650,000 people and at the end of 25 years that number would be 9,125,000.

The disciplemakers keep on investing in one new person every year who does the same thing, he invests in one new person every year. The numbers simply multiply and at the end of 10 years there are 1,024 disciplemakers and at the end of 25 years, there would be 33,554,432 disciplemakers (over 3x more than the addition process).


It’s easy to see that the process of multiplication is slower than the process of addition. It takes 19 years for the number of disciplemakers to exceed the number of the first year alone. However, when the disciplemaking process reaches year 26, they would reach 67,108,864, a number you couldn’t reach by addition for another 158 years.

The big difference in disciplemaking isn’t just the numbers. It’s the growth and maturity that happen in the disciplemaking process! When you focus on disciplemaking, you develop men and women who are mature in their faith and able to spiritually reproduce themselves in others.

Today’s Missional Challenge

Don’t just add new believers. Multiply disciplemakers. Make disciples who make disciples who make disciples who make disciples…


8.2: Addition and Multiplication Principles

We often speak of `` the field " instead of `` the field ".

The right side of (2.51) is ambiguous. There are five sensible ways to interpret it:

  1. Multiplication and division have equal precedence.
  2. Addition and subtraction have equal precedence.
  3. Multiplication has higher precedence than addition.

you first read (2.52) from left to right and perform all the multipliations and divisions as you come to them, getting

Then read (2.53) from left to right performing all additions and subtractions as you come to them, getting

When I was in high school, multiplication had higher precedence than division, so

In 1713, addition often had higher precedence than multiplication. Jacob Bernoulli [8, p180] wrote expressions like

is a field. (See definition 2.42 for the definitions.) We showed in section 2.2 that satisfies all the field axioms except possibly the distributive law. In appendix B, it is shown that the distributive property holds for for all , . (The proof assumes that the distributive law holds in .)

For a general , , the only field axiom that can possibly fail to hold in is the existence of multiplicative inverses, so to determine whether is a field, it is just necessary to determine whether every non-zero element in is invertible for .


Multi-Digit Multiplication

On this page you have a large selection of 2-digit by 1-digit multiplication worksheets to choose from. (example: 32x5)

On these PDF files, students can find the products of 3-digit numbers and 1-digit numbers. (example: 371x3)

Review 4-digit by 1-digit multiplication problems with these worksheets and task cards. (example: 3,812x7)

Here's a link to a set of worksheets with 2-digit by 2-digit multiplication problems on them. Includes math riddles, a Scoot game, task cards, and more. (example: 43x19)

On these printables, your pupils will be multiplying 3-digit numbers by 2-digit numbers. (example: 778x2)

This collection features worksheets that require students to multiply by 3-digit numbers. (example: 235x129)

This page has lots of worksheets on finding the products of pairs of decimal numbers. (example: 1.3x5.6)

These worksheets will have students multiplying money amounts. (example: $5.67x3)


Watch the video: Go Math 5th Grade Lesson Problem Solving Use Multiplication UPDATED (November 2021).