In many counting problems, the order of arrangement or selection does not matter. In essence, we are selecting or forming subsets.

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

Determine the number of ways to choose 4 values from 1, 2, 3, …, 20, in which the order of selection does not matter.

**Solution**Let (N) be the number of ways to choose the 4 numbers. Since the order in which the numbers are selected does not matter, these are

*not*sequences (in which order of appearance matters). We can change a selection of 4 numbers into a sequence. The 4 numbers can be arranged in (P(4,4)=4!) ways. Therefore, all these 4-number selections together produce (Ncdot4!) sequences. The number of 4-number sequences is (P(20,4)). Thus, (Ncdot4!=P(20,4)), or equivalently, (N=P(20,4)/4!).

Definition: combinations

The number of (r)-element subsets in an (n)-element set is denoted by

[C(n,r) qquadmbox{ or }qquad inom{n}{r}, onumber]

where ({nchoose r}) is read as “(n) choose (r).” It determines the number of ** combinations** of (n) objects, taken (r) at a time (without replacement). Alternate notations such as (_nC_r) and (C_r^n) can be found in other textbooks. Do

*not*write it as ((frac{n}{r})); this notation has a completely different meaning.

Recall that (inom{n}{r}) counts the number of ways to *choose* or *select* (r) objects from a pool of (n) objects in which the order of selection does not matter. Hence, (r)-combinations are subsets of size (r).

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

The 2-combinations of (S={a,b,c,d}) are

[{a,b}, quad {a,c}, quad {a,d}, quad {b,c}, quad {b,d}, quadmbox{and}quad {c,d}. onumber]

Therefore (inom{4}{2}=6). What are the 1-combinations and 3-combinations of (S)? What can you say about the values of (inom{4}{1}) and (inom{4}{3})?

**Solution**The 1-combinations are the singleton sets ({a}), ({b}), ({c}), and ({d}). Hence, (inom{4}{1}=4). The 3-combinations are [{a,b,c}, quad {a,b,d}, quad {a,c,d}, quadmbox{and}quad {b,c,d}. onumber] Thus, (inom{4}{3}=4).

Theorem (PageIndex{1}label{thm:combin})

For all integers (n) and (r) satisfying (0leq rleq n), we have [inom{n}{r} = frac{P(n,r)}{r!} = frac{n(n-1)cdots(n-r+1)}{r!} = frac{n!}{r!,(n-r)!}. onumber]

**Proof**The idea is similar to the one we used in the alternate proof of Theorem 8.3.2. Let (A) be the set of all (r)-permutations, and let (B) be the set of all (r)-combinations. Define (f: A o B) to be the function that converts a permutation into a combination by “unscrambling” its order. Then (f) is an (r!)-to-one function because there are (r!) ways to arrange (or shuffle) (r) objects. Therefore [|A| = r!cdot|B|. onumber] Since (|A|=P(n,r)), and (|B|=inom{n}{r}), it follows that (inom{n}{r} = P(n,r)/r!).

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

There are (inom{40}{5}) ways to choose 5 numbers, without repetitions, from the integers (1,2,ldots,40). To compute its numeric value by hand, it is easier if we first cancel the common factors in the numerator and the denominator. We find

[inom{40}{5} = frac{40cdot39cdot38cdot37cdot36} {5cdot4cdot3cdot2cdot1} = 13cdot38cdot37cdot36, onumber]

which gives (inom{40}{5}=658008).

hands-on Exercise (PageIndex{1}label{he:combin-01})

Compute (inom{12}{3}) by hand.

hands-on Exercise (PageIndex{2}label{he:combin-02})

A three-member executive committee is to be selected from a group of seven candidates. In how many ways can the committee be formed?

hands-on Exercise (PageIndex{3}label{he:combin-03})

How many subsets of ({1,2,ldots,23}) have five elements?

Corollary (PageIndex{2})

For (0leq rleq n), we have (inom{n}{r} = inom{n}{n-r}).

**Proof**According to Theorem 8.4.1, we have [inom{n}{n-r} = frac{n!}{(n-r)!,(n-(n-r))!} = frac{n!}{(n-r)!,r!}, onumber] which is precisely (inom{n}{r}).

Example (PageIndex{1}label{eg:combin-04})

To compute the numeric value of (inom{50}{47}), instead of computing the product of 47 factors as indicated in the definition, it is much faster if we use [inom{50}{47} = inom{50}{3} = frac{50cdot 49cdot48}{3cdot 2cdot 1}, onumber] from which we obtain (inom{50}{47} = 19600).

hands-on Exercise (PageIndex{4}label{he:combin-04})

Compute, by hand, the numeric value of (inom{529}{525}).

Now we are ready to look at some mixed examples. In all of these examples, sometimes we have to use permutation, other times we have to use combination. Very often we need to use both, together with the addition and multiplication principles. You may ask, how can I figure out what to do? We suggest asking yourself these questions:

- Use the construction approach. If you want to list all the configurations that meet the requirement, how are you going to do it systematically?
- Are there several cases involved in the problem? If yes, we need to list them first,
*before*we go through each of them one at a time. Finally, add the results to come up with the final answer. - Do we allow repetitions or replacements? This question can also take the form of whether the objects are distinguishable or indistinguishable.
- Does order matter? If yes, we have to use permutation. Otherwise, use combination.
- Sometimes, it may be easier to use the multiplication principle instead of permutation, because repetitions may be allowed (in which case, we cannot use permutation, although we can still use the multiplication principle). Try drawing a schematic diagram and decide what we need from it. If the analysis suggests a pattern that follows the one found in a permutation, you can then use the formula for permutation.
- Do not forget: it may be easier to work with the complement.

It is often not clear how to get started because there seem to be several ways to start the construction. For example, how would you distribute soda cans among a group of students? There are two possible approaches:

- From the perspective of the students. Imagine you are one of the students, which soda would you receive?
- From the perspective of the soda cans. Imagine you are holding a can of soda, to whom would you give this soda?

Depending on the actual problem, usually only one of these two approaches would work.

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

Suppose we have to distribute 10 different soda cans to 20 students. It is clear that some students may not get any soda. In fact, some lucky students could receive more than one soda (the problem does not say this cannot happen). Hence, it is easier to start from the perspective of the soda cans.

**Solution**We can give the first soda to any one of the 20 students, and we can also give the second soda to any one of the 20 students. In fact, we always have 20 choices for each soda. Since we have 10 sodas, there are (underbrace{20cdot20cdots20}_{10} = 20^{10}) ways to distribute the sodas.

Example (PageIndex{6}label{eg:combin-06})

In how many ways can a team of three representatives be selected from a class of 885 students? In how many ways can a team of three representatives consisting of a chairperson, a vice-chairperson, and a secretary be selected?

**Solution**If we are only interested in selecting three representatives, order does not matter. Hence, the answer would be (inom{885}{3}). If we are concerned about which offices these three representative will hold, then the answer should be (P(885,3)).

hands-on Exercise (PageIndex{5}label{he:combin-05})

Mike needs some new shirts, but he has only enough money to purchase five of the eight that he likes. In how many ways can he purchase the five shirts by choosing them at random?

hands-on Exercise (PageIndex{6}label{he:combin-06})

Mary wants to purchase four shirts for her four brothers, and she would like each of them to receive a different shirt. She finds ten shirts that she thinks they will like. In many ways can she select them?

Playing cards provide excellent examples for counting problems. Just in case you are not familiar with them, let us briefly review what a deck of playing cards contains.

- There are 52 playing cards, each of them is marked with a suit and a rank.
- There are four suits: spades ((spadesuit)), hearts ((heartsuit)), diamonds ((diamondsuit)) and clubs ((clubsuit)).
- Each suit has 13 ranks, labeled A, 2, 3, …, 9, 10, J, Q, and K, where A means ace, J means jack, Q means queen, and K means king.
- Each rank has 4 suits (see above).

hands-on Exercise (PageIndex{7}label{he:combin-07})

Determine the number of five-card poker hands that can be dealt from a deck of 52 cards.

**Solution**All we care is which five cards can be found in a hand. This is a selection problem. The answer is (inom{52}{5}).

hands-on exercise (PageIndex{7}label{eg:combin-07})

In how many ways can a 13-card bridge hand be dealt from a standard deck of 52 cards?

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

In how many ways can a deck of 52 cards be dealt in a game of bridge? (In a bridge game, there are four players designated as North, East, South and West, each of them is dealt a hand of 13 cards.)

**Solution**The difference between this problem and the last example is that the order of distributing the four bridge hands makes a difference. This is a problem that combines permutations and combinations. As we had suggested earlier, the best approach is to start from scratch, using the addition and/or multiplication principles, along with permutation and/or combination whenever it seems appropriate.

There are (inom{52}{13}) ways to give 13 cards to the first player. Now we are left with 39 cards, from which we select 13 to be given to the second player. Now, out of the remaining 26 cards, we have to give 13 to the third player. Finally, the last 13 cards will be given to the last player (there is only one way to do it). The number of ways to deal the cards in a bridge game is (inom{52}{13} inom{39}{13} inom{26}{13}).

We could have said the answer is [inom{52}{13} inom{39}{13} inom{26}{13} inom{13}{13}. onumber] The last factor (inom{13}{13}) is the number of ways to give the last 13 cards to the fourth player. Numerically, (inom{13}{13}=1), so the two answers are the same. Do not dismiss this extra factor as redundant. Take note of the nice pattern in this answer. The bottom numbers are 13, because we are selecting 13 cards to be given to each player. The top numbers indicate how many cards are still available for distribution at each stage of the distribution. The reasoning behind the solution is self-explanatory!

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

Determine the number of five-card poker hands that contain three queens. How many of them contain, in addition to the three queens, another pair of cards?

**Solution**- The first step is to choose the three queens in (inom{4}{3}) ways, after which the remaining two cards can be selected in (inom{48}{2}) ways. Therefore, there are altogether (inom{4}{3} inom{48}{2}) hands that meet the requirements.
- As in part (a), the three queens can be selected in (inom{4}{3}) ways. Next, we need to select the pair. We can select any card from the remaining 48 cards (therefore, there are 48 choices), after which we have to select one from the remaining 3 cards of the same rank. This gives (48cdot3) choices for the pair, right? The answer is
*NO*!

The first card we picked could be (heartsuit 8), and the second could be (clubsuit 8). However, the first card could have been (clubsuit 8), and the second (heartsuit 8). These two selections are counted as

*different*selections, but they are actually the same pair! The trouble is, we are considering “first,” and “second” cards, which in effect imposes an ordering among the two cards, thereby turning it into a sequence or an*ordered*selection. We have to divide the answer by 2 to overcome the double-counting. The answer is therefore (frac{48cdot3}{2}).Here is a better way to count the number of pairs. An important question to ask is

*Which one should we pick first: the suit or the rank?*Here, we want to pick the rank first. There are 12 choices (the pair cannot be queens) for the rank, and among the four cards of that rank, we can pick the two cards in (inom{4}{2}) ways. Therefore, the answer is (12inom{4}{2}). Numerically, the two answers are identical, because (12inom{4}{2} = 12cdotfrac{4cdot3}{2} = frac{48cdot3}{2}). In summary: the final answer is (inom{4}{3}cdot12inom{4}{2}).

hands-on Exercise (PageIndex{8}label{he:combin-08})

How many bridge hands contain exactly four spades?

hands-on Exercise (PageIndex{9}label{he:combin-09})

How many bridge hands contain exactly four spades and four hearts?

hands-on Exercise (PageIndex{10}label{he:combin-10})

How many bridge hands are there containing exactly four spades, three hearts, three diamonds, and three clubs?

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

How many positive integers not exceeding 99999 contain exactly three 7s?

**Solution**Regard each legitimate integer as a sequence of five digits, each of them selected from 0, 1, 2, …, 9. For example, the integer 358 can be considered as 00358. Three out of the five positions must be occupied by 7. There are (inom{5}{3}) ways to select these three slots. The remaining two positions can be filled with any of the other nine digits. Hence, there are (inom{5}{3} cdot 9^2) such integers.

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

How many five-digit positive integers contain exactly three 7s?

**Solution**Unlike the last example, the first of the five digits cannot be 0. Yet, the answer is

*not*(inom{5}{3} cdot 9 cdot 8). Yes, there are (inom{5}{3}) choices for the placement of the three 7s, but some of these selections may have put the 7s in the last four positions. This leaves the first digit unfilled. The nine choices counted by 9 allows a zero to be placed in the first position. The result is, at best, a four-digit number. The correct approach is to consider two cases:- Case 1. If the first digit is not 7, then there are eight ways to fill this slot. Among the remaining four positions, three of them must be 7, and the last one can be any digit other than 7. So there are (8cdot inom{4}{3}cdot 9) integers in this category.
- Case 2. If the first digit is 7, we still have to put the other two 7s in the other four positions. There are (inom{4}{2} cdot 9^2) such integers.

Together, the two cases give a total of (8cdot inom{4}{3}cdot 9 + inom{4}{2} cdot 9^2 = 774) integers.

hands-on Exercise (PageIndex{11}label{he:combin-11})

Five balls are chosen from a bag of eight blue balls, six red balls, and five green balls. How many of these five-ball selections contain exactly two blue balls?

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

Find the number of ways to select five balls from a bag of six red balls, eight blue balls and four yellow balls such that the five-ball selections contain exactly two red balls *or* two blue balls.

**Solution**The keyword “or” suggests this is a problem that involves the union of two sets, hence, we have to use PIE to solve the problem.

- How many selections contain two red balls? Following the same argument used in the last example, the answer is (inom{6}{2} inom{12}{3}).
- How many selections contain two blue balls? The answer is (inom{8}{2} inom{10}{3}).
- According to PIE, the final answer is [inom{6}{2} inom{12}{3} + inom{8}{2} inom{10}{3} - inom{6}{2} inom{8}{2} inom{4}{1}. onumber] In each term, the upper numbers always add up to 18, and the sum of the lower numbers is always 5. Can you explain why?
- How many selections contain two red balls
*and*2 blue balls? The answer is (inom{6}{2} inom{8}{2} inom{4}{1}).

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

We have 11 balls, five of which are blue, three of which are red, and the remaining three are green. How many collection of four balls can be selected such that at least two blue balls are selected? Assume that balls of the same color are indistinguishable.

**Solution**The keywords “at least” mean we could have two, three, or four blue balls. There are [inom{5}{2} inom{6}{2} + inom{5}{3} inom{6}{1} + inom{5}{4} inom{6}{0} onumber] ways to select four balls, with at least two of them being blue.

hands-on Exercise (PageIndex{12}label{he:combin-12})

Jerry bought eight cans of Pepsi, seven cans of Sprite, three cans of Dr. Pepper, and six cans of Mountain Dew. He want to bring 10 cans to his pal’s house when they watch the basketball game tonight. Assuming the cans are distinguishable, say, with different expiration dates, how many selections can he make if he wants to bring

- Exactly four cans of Pepsi?
- At least four cans of Pepsi?
- At most four cans of Pepsi?
- Exactly three cans of Pepsi, and at most three cans of Sprite?

The proof of the next result uses what we call a combinatorial or counting argument. In general, a combinatorial argument does not rely on algebraic manipulation. Rather, it uses the combinatorial significance of the situations to solve the problem.

Theorem (PageIndex{3})

Prove that (sum_{r=0}^n inom{n}{r} = 2^n) for all nonnegative integers (n).

**Proof**Since (inom{n}{r}) counts the number of (r)-element subsets selected from an (n)-element set (S), the summation on the left is the sum of the number of subsets of (S) of all possible cardinalities. In other words, this is the total number of subsets in (S). We learned earlier that (S) has (2^n) subsets, which establishes the identity immediately.

## Summary and Review

- Use permutation if order matters, otherwise use combination.
- The keywords arrangement, sequence, and order suggest using permutation.
- The keywords selection, subset, and group suggest using combination.
- It is best to start with a construction. Imagine you want to list all the possibilities, how would you get started?
- We may need to use both permutation and combination, and very likely we may also need to use the addition and multiplication principles.

Exercise (PageIndex{1}label{ex:combin-01})

If the Buffalo Bills and the Cleveland Browns have eight and six players, respectively, available for trading, in how many ways can they swap three players for three players?

Exercise (PageIndex{2}label{ex:combin-02})

In the game of Mastermind, one player, the codemaker, selects a sequence of four colors (the “code”) selected from red, blue, green, white, black, and yellow.

- How many different codes can be formed?
- How many codes use four different colors?
- How many codes use only one color?
- How many codes use exactly two colors?
- How many codes use exactly three colors?

Exercise (PageIndex{3}label{ex:combin-03})

Becky likes to watch DVDs each evening. How many DVDs must she have if she is able to watch every evening for 24 consecutive evenings during her winter break?

- A different subset of DVDs?
- A different subset of three DVDs?

Exercise (PageIndex{4}label{ex:combin-04})

Bridget has (n) friends from her bridge club. Every Thursday evening, she invites three friends to her home for a bridge game. She always sits in the north position, and she decides which friends are to sit in the east, south, and west positions. She is able to do this for 200 weeks without repeating a seating arrangement. What is the minimum value of (n)?

Exercise (PageIndex{5}label{ex:combin-05})

Bridget has (n) friends from her bridge club. She is able to invite a different subset of three of them to her home every Thursday evening for 100 weeks. What is the minimum value of (n)?

Exercise (PageIndex{6}label{ex:combin-06})

How many five-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6, 7? How many of them do not have repeated digits?

Exercise (PageIndex{7}label{ex:combin-07})

The Mathematics Department of a small college has three full professors, seven associate professors, and four assistant professors. In how many ways can a four-member committee be formed under these restrictions:

- There are no restrictions.
- At least one full professor is selected.
- The committee must contain a professor from each rank.

Exercise (PageIndex{8}label{ex:combin-08})

A department store manager receives from the company headquarters 12 football tickets to the same game (hence they can be regarded as “identical”). In how many ways can she distribute them to 20 employees if no one gets more than one ticket? What if the tickets are for 12 different games?

Exercise (PageIndex{9}label{ex:combin-09})

A checkerboard has 64 distinct squares arranged into eight rows and eight columns.

- In how many ways can eight identical checkers be placed on the board so that no two checkers can occupy the same row or the same column?
- In how many ways can two identical red checkers and two identical black checkers be placed on the board so that no two checkers of the same color can occupy the same row or the same column?

Exercise (PageIndex{10}label{ex:combin-10})

Determine the number of permutations of ({A, B, C, D, E}) that satisfy the following conditions:

- (A) occupies the first position.
- (A) occupies the first position, and (B) the second.
- (A) appears before (B).

Exercise (PageIndex{11}label{ex:combin-11})

A binary string is a sequence of digits chosen from 0 and 1. How many binary strings of length 16 contain exactly seven 1s?

Exercise (PageIndex{12}label{ex:combin-12})

In how many ways can a nonempty subset of people be chosen from eight men and eight women so that every subset contains an equal number of men and women?

Exercise (PageIndex{13}label{ex:combin-13})

A poker hand is a five-card selection chosen from a standard deck of 52 cards. How many poker hands satisfy the following conditions?

- There are no restrictions.
- The hand contains at least one card from each suit.
- The hand contains exactly one pair (the other three cards all of different ranks).
- The hand contains three of a rank (the other two cards all of different ranks).
- The hand is a full house (three of one rank and a pair of another).
- The hand is a straight (consecutive ranks, as in 5, 6, 7, 8, 9, but not all from the same suit).
- The hand is a flush (all the same suit, but not a straight).
- The hand is a straight flush (both straight and flush).

Exercise (PageIndex{14}label{ex:combin-14})

A local pizza restaurant offers the following toppings on their cheese pizzas: extra cheese, pepperoni, mushrooms, green peppers, onions, sausage, ham, and anchovies.

- How many kinds of pizzas can one order?
- How many kinds of pizzas can one order with exactly three toppings?
- How many kinds of vegetarian pizza (without pepperoni, sausage, or ham) can one order?

## Combinations - Calculating the outcomes for a 15-coin collection

I searched a lot to find some sort of guidance on how to approach solving this problem, but I couldn't find an answer anywhere. Read the book many times, still no luck The problem as follows:

Cecil has a 15-coin collection. Four coins are quarters, seven coins are dimes, three are nickels and one is a penny. For each scenario, calculate the total possible outcomes if Cecil randomly selects five coins.

Are my answers to each scenario correct? If not, what am I missing? Thanks in advance for any answers.

In how many different ways can we select 3 items from a set of 8?

#### Number of k-combinations

The number of *k*-combinations from a given set *S* of *n* elements can be denoted by

### Calculating the number of combinations

Using the easy to remember symmetric formula

Substituting our values for n=8 and k=3 we get

###### Expending factorial and cancelling out

#### Final Result :56

There are 56 (fifty-six ) combinations to choose 3 items out of a set of 8

A **system of equations** is a set of 2 (or more) equations where the variables represent the same unknown values. For example, suppose that two different kinds of bamboo are planted at the same time. Plant A starts at 6 ft tall and grows at a constant rate of frac14 foot each day. Plant B starts at 3 ft tall and grows at a constant rate of frac12 foot each day. We can write equations y = frac14 x + 6 for Plant A and y = frac12 x +3 for Plant B, where x represents the number of days after being planted, and y represents height. We can write this system of equations.

egin

Solving a system of equations means to find the values of x and y that make both equations true at the same time. One way we have seen to find the solution to a system of equations is to graph both lines and find the intersection point. The intersection point represents the pair of x and y values that make both equations true. Here is a graph for the bamboo example:

The solution to this system of equations is (12,9) , which means that both bamboo plants will be 9 feet tall after 12 days.

We have seen systems of equations that have no solutions, one solution, and infinitely many solutions.

- When the lines do not intersect, there is no solution. (Lines that do not intersect are
*parallel*.) - When the lines intersect once, there is one solution.
- When the lines are right on top of each other, there are infinitely many solutions.

In future lessons, we will see that some systems cannot be easily solved by graphing, but can be easily solved using algebra.

## Binomial Theorem

- You have often has to expand polynomials like this: [egin
(x+y)^4 &= (x+y)(x+y)(x+y)(x+y) &= (x^2+2xy+y^2)(x+y)(x+y) &= (x^3+3x^2y+3xy^2+y^3)(x+y) &= x^4+4x^3y+6x^2y^2+4xy^3+y^4 end ] - &hellip and it was a little painful.

*Theorem:*For a non-negative integer (n), [ (x+y)^n = sum_*^*inom *x^*y^i,. ] Recall that (inom *) is another notation for (C(n,i)).**Proof idea:*The coefficient of the (x^y^i) term comes from the number ways that it is possible to choose the first term (n-i) time and the second term (i) times while doing the expansion. We have (n) terms to work with, and must select (i) of them to multiply the (y). There are (inom *) ways to do this.**Corollary:*For a non-negative integer (n), [ sum_*^*inom *=2^n ,. ]**Proof:*Apply the binomial theorem with (x=y=1). ∎- The left side of the equation is the number of bit strings of length (n) with 0 ones, plus bit strings with 1 one, with 2 ones, &hellip, (n) ones.
- The right side is the number of bit strings of length (n).
- Those count the same thing, so they must be equal.

## How to handle GMAT Math questions on Combinations and Probability

The GMAT quantitative (Maths) section often has questions involving concepts of Combinations and/or Probability. The GMAT preparation series on MBA Crystal Ball moves on, as the GoGMAT team covers questions that cover

*both*Combinations and Probability .### GMAT Maths: Combinations and Probability

The GMAT exam is designed to surprise you. There are no books detailing all the exact questions you are likely to face, and there is nobody who can walk into a test center with full confidence of scoring 800.

Some GMAT math questions are straightforward, and test a single math concept. However, harder questions, and those are the ones that contribute the most to your score, tend to mix different math topics. Questions that mix Combinations and Probability are a great example. There are almost unlimited examples however, once you learn the core principles, you should be able to understand quickly what the question requires.

Unfortunately, these types of questions also seem to be the wordiest. We have to learn to step back from the wording and identify the important information fast. Thankfully this is easier then it looks when you have good methodology.

Let’s begin by reminding ourselves of the formulas for probability and combinations/permutations.

The formula for simple probability:

Probability of A happening = (number of times A can occur) / ( number of time

**any**outcome can occur)The formula for Permutations:

And the formula for Combinations:

Remember: in Permutations, the order matters. In Combinations—it does not.

**Methodology:**There are two steps that will help you deal with GMAT probability and combinations/permutations questions efficiently.

1) Work out whether the question asks for Combinations or Permutations

2) Write out the equation and fill in what’s needed.

You simply dig into the wording and take out the information necessary to get to the answer fast.

*Two dice are rolled. What is the probability that the sum will be greater than 10?*At the first glance, we can see that this question involves probability and calculating the number of ways in which any number can be rolled. So let’s look at the first step in our methodology:

*1)**Work out whether the question asks for Combinations or Permutations*In this question the order of the numbers rolled is important. If we roll a 6 on die one, and a 4 on die two, then that is different from a 4 on die one and a 6 on die two they are two different results, which naturally affects our probability calculation. Therefore, this question deals with permutations.

Working out the permutations of numerous similar objects such as dice is straightforward and does not require the use of a formula.

There are six different possibilities on a single die, as we can roll either a 1, 2, 3, 4, 5 or 6. If we have two dice, then the number of permutations is simply:

6 possible outcomes on die one X 6 possible outcomes on die two

6 X 6 = 36 total permutations

Now we can move to step 2:

*2)**Write out the equation and fill in what’s needed.*Our probability equation will look like:

Probability of rolling greater than 10 = (number of outcomes over 10) / ( total number of outcomes )

So far, we only know the denominator as we worked out the total number of permutations in step one. We now need to work out the number of times we can roll a number over ten. This step requires some basic maths and logical thinking. If we need to roll over ten with just two dice, then the number rolled must be either 11 or 12. So now, we need to calculate the number of ways in which we can roll 11 or 12.

We can do this the long way by calculating every permutation and it’s value:

## 8.4 Expression of a Vector as the Linear Combination of a Few Vectors

2.**prove that two vectors**are**parallel**, we must express one of the vectors as a**scalar multiple**of the other vector.

3.**points***P*,*Q*and*R*are**collinear**, prove one of the following.• P Q → = k Q R → or Q R → = h P Q → • P R → = k P Q → or P Q → = h P R → • P R → = k Q R → or Q R → = h P R →

(a)(i) A S → = A D → + D S → = A D → + A Q → ← A Q : Q B = 3 : 1 and D S : S C = 3 : 1 ∴ A Q → = D S → = b ˜ + 6 a ˜ = 6 a ˜ + b ˜

(a)(ii) Q C → = Q B → + B C → = 1 3 A Q → + A D → ← A Q : Q B = 3 : 1 A Q Q B = 3 1 ⇒ Q B = 1 3 A Q and for parallelogram, B C / / A D , B C = A D = 1 3 ( 6 a ˜ ) + b ˜ = 2 a ˜ + b ˜

(b) Q T → = Q A → + A T → = Q A → + 3 2 A S → ← A S = 2 S T A T = 3 S T = 3 2 A S = − 6 a ˜ + 3 2 ( 6 a ˜ + b ˜ ) = 3 a ˜ + 3 2 b ˜ = 3 2 ( 2 a ˜ + b ˜ ) = 3 2 Q C → ∴ Points Q , C and T are collinear .

## 2018 April(A09) Math Question 36

**You are not logged in.**Log In | Sign Up and pay $59 now to view all explanations.

## Combinations without repetition

The combinations without repetition of $n$ elements taken $k$ in $k$ are the different groups of $k$ elements that can be formed by these $n$ elements, so that two groups differ only if they have different elements (that is to say, the order does not matter). They are represented as $C_

$. Let's consider the set $A=$ of $5$ elements. Let's observe first of all that, for example, the groups $abc$ and $cba$ are considered to be equal, since as has been said the order does not matter while the elements are the same.

We are going to see what the different combinations without repetition of these $5$ elements are:

- Combinations without repetition of $5$ elements taken $1$ at a time: $a$, $b$, $c$, $d$ and $e$.
- Combinations without repetition of $5$ elements taken $2$ at a time: $ab$, $ac$, $ad$, $ae$, $bc$, $bd$, $be$, $cd$, $ce$ and $de$.
- Combinations without repetition of $5$ elements taken $3$ at a time: $abc$, $abd$, $abe$, $acd$, $ace$, $ade$, $bcd$, $bce$, $bde$ and $cde$.
- Combinations without repetition of $5$ elements taken $4$ at a time: $abcd$, $abce$, $abde$, $acde$ and $bcde$.
- Combinations without repetition of $5$ elements taken $5$ at a time: The only group of $5$ elements that it is possible to form from the elements of $A$ is $abcde$.

In this example all of the combinations could have been written. However, if $A$ had had many more elements, this would have been much more complicated.

The following formula allows us to know how many combinations without repetition of $n$ elements taken $k$ in $k$ there are: $$displaystyle C_

=inom = frac $$ In the previous example, $n = 5$. Now, if we want to know how many combinations of $5$ elements, taken $3$ at a time there are, we use the formula and we obtain: $$displaystyle C_<5,3>=inom<5> <3>= frac<5!><3!(5-3)!>=10$$ We can check in the previous list that there are $10$ sets of $3$ elements, indeed.

## Features

**Extend students’ mathematical maturity and ability to deal with abstraction****Strong emphasis on reading and writing proofs**– Illustrates most proofs of theorems with annotated figures to provide additional explanation and insight into the proofs.**EXPANDED! More than 100 new exercises**have been added to the first three chapters: Sets and Logic, Proofs, and Functions, Sequences, and Relations. There are now more than 1,750 worked examples and exercises in these chapters.**Problem Solving Corners**, a hallmark feature that helps students attack and solve problems and show them how to do proofs.

**Breadth of examples and exercises help students master introductory discrete mathematics**

## Watch the video: Section Permutation u0026 Combinations (December 2021).