# 7.3: Permutations and Combinations - Mathematics

We saw in the last section that, when working with permutations, the order is always important. If we were choosing 3 people from a group of 7 to serve on a committee with no assigned roles, the nature of the problem would change.
For example, if we were choosing 3 people from a group of 7 to serve on a committee as president, vice-president and treasurer, the answer would be (_{7} P_{3}=210) But - if we wanted to choose 3 people from a group of 7 with no assigned roles, then some of the choices in the permutation would be the same.
In a permutation:
1st place: Alice 1st place: Bob 2nd place: Bob (quad) 2nd place: Charlie 3rd place: Charlie (quad) 3rd place: Alice
the two choices listed above would be considered as being different and would be counted separately. In a "combination" in which the order of selection is not important and there are no assigned roles, we must compensate for these extra choices.

If we are choosing 3 people from a group of 7 to serve on a committee with no assigned roles then we should consider that any selection from a permutation that includes the same three people should only be counted once.
So, when we select the three people, we should consider how many different ways there are to group them and then remove those extra choices. In this example, we are choosing three people. Each group of three can be arranged in six different ways (3 !=3 * 2=6,) so each distinct group of three is counted six times.
In order to find the actual number of choices we take the number of possible permutations and divide by 6 to arrive at the actual answer:
[
_{7} C_{3}=frac{7 P_{3}}{3 !}=frac{7 !}{4 ! * 3 !}
]
In a combination in which the order is not important and there are no assigned roles the number of possibilities is defined as:
[
_{n} C_{r}=frac{n !}{(n-r) ! * r !}
]
One way to remember the difference between a permutation and a combination is that on a combination pizza it doesn't make any difference whether the sausage goes on before the pepperoni or whether the onions are put on first-so in a combination, order is not important!

EXERCISES 7.3
Find the value of the following expressions.
9) How many three-topping pizzas can be made if there are twelve toppings to choose from?
10) How many bridge hands of 13 cards are possible from a deck of 52 cards?
11) How many poker hands of 5 cards are possible from a deck of 52 cards?
12) How many different bridge hands of 13 cards are possible if none of the cards is higher than 10 (i.e. no face cards)?
13) How many different poker hands of 5 cards are possible if none of the cards is higher than (8 ?)
14) If a person has 10 different t-shirts, how many ways are there to choose 4 to take on a trip?
15) If a band has practiced 15 songs, how many ways are there for them to select 4 songs to play at a battle of the bands? How many different performances of four songs are possible?
16) Fifteen boys and 12 girls are on a camping trip. How many ways can a group of seven be selected to gather firewood:
(quad) b) the group contains four girls and three boys
(quad) c) the group contains at least four girls
17) A class of 25 students is comprised of 15 girls and 10 boys. In how many ways can a committee of 8 students be selected if:
(quad) a) there are no restrictions
(quad) b) no males are included on the committee
(quad) c) no females are included on the committee
(quad) d) the committee must have 5 boys and 3 girls
18) From a group of 12 male and 12 female tennis players, two men and two women will be chosen to compete in a men-vs-women doubles match. How many different matches are possible?
19) In a seventh-grade dance class, there are 20 girls and 17 boys.
(quad) a) How many ways can the students be paired off to create dance couples consisting of one boy and one girl?
(quad) b) How many ways are there to create a group of 17 boy/girl couples?
(quad) c) How many ways are there to create a group of 18 couples without restrictions?

## NCERT Solutions for Class 11 Maths Chapter 7 Permutation and Combinations

Permutations and Combinations Class 11 Maths NCERT Solutions are extremely helpful while doing your homework. NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations All Exercises were prepared by Experienced LearnCBSE.in Teachers.

Free download NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.1, Ex 7.2, Ex 7.3 , Ex 7.4 and Miscellaneous Exercise PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

## Permutation and Combination – Class XI – Exercise 7.3

3-digit numbers have to be formed using the digits 1 to 9. Here, the order of the digits matters.

Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.

Therefore, required number of 3-digit numbers = 9 P3 = 9! /(9-3)! = 9! /6!

2: How many 4-digit numbers are there with no digit repeated?

The thousands place of the 4-digit number is to be filled with any of the digits from 1 to 9 as the digit 0 cannot be included.

Therefore, the number of ways in which thousands place can be filled is 9.

The hundreds, tens, and units place can be filled by any of the digits from 0 to 9.

However, the digits cannot be repeated in the 4-digit numbers and thousands place is already occupied with a digit.

The hundreds, tens, and units place is to be filled by the remaining 9 digits.

Therefore, there will be as many such 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.

Number of such 3 – digit numbers = 9 P3 = 9! /(9-3)! = 9! /6!

Thus, by multiplication principle, the required number of 4-digit numbers is 9 × 504 = 4536

1. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

3-digit even numbers are to be formed using the given six digits, 1, 2, 3, 4, 6, and 7, without repeating the digits.

Then, units digits can be filled in 3 ways by any of the digits, 2, 4, or 6. Since the digits cannot be repeated in the 3-digit numbers and units place is already occupied with a digit (which is even), the hundreds and tens place is to be filled by the remaining 5 digits.

Therefore, the number of ways in which hundreds and tens place can be filled with the remaining 5 digits is the permutation of 5 different digits taken 2 at a time.

Number of ways of filling hundreds and tens place = 5 P2 = 5! /(5-2)! = 5! /3!

Thus, bu multiplication principle, the required number of 3 digit numbers 3 x 20 = 60

4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

4-digit numbers are to be formed using the digits, 1, 2, 3, 4, and 5.

There will be as many 4-digit numbers as there are permutations of 5 different digits taken 4 at a time.

Therefore, required number of 4 digit numbers = 5 P4 = 5! /(5-4)! = 5! /1!

Among the 4-digit numbers formed by using the digits, 1, 2, 3, 4, 5, even numbers end with either 2 or 4.

The number of ways in which units place is filled with digits is 2. Since the digits are not repeated and the units place is already occupied with a digit (which is even), the remaining places are to be filled by the remaining 4 digits.

Therefore, the number of ways in which the remaining places can be filled is the permutation of 4 different digits taken 3 at a time.

Number of ways of filling the remaining places = 4 P3 = 4! /(4-3)! = 4! /1!

Thus, by multiplication principle, the required number of even numbers is 24 × 2 = 48

1. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?

From a committee of 8 persons, a chairman and a vice chairman are to be chosen in such a way that one person cannot hold more than one position.

Here, the number of ways of choosing a chairman and a vice chairman is the permutation of 8 different objects taken 2 at a time.

Thus, required number of ways = 8 P2 = 8! /(8-2)! = 8! /6! = 8x7x6! /6! = 8 x 7 = 56

(i) 5 Pr = 2 6 Pr-1

1. How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

We have 8 different letters in the word EQUATION.

Therefore, the number of words that can be formed using all the letters of the word EQUATION, using every letter only once, is the number of permutations of 8 different objects taken 8 at a time, which is 8 P8 = 8!

Thus, the required number of words that can be formed = 8! = 40320

9: How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if

(i) 4 letters are used at a time,

(ii) all letters are used at a time,

(iii) all letters are used but first letter is a vowel?

There are 6 different letters in the word MONDAY.

(i) Number of 4-letter words can be formed from the letters of the word MONDAY without repetition of letters is equal to the number of permutations of 6 different objects taken 4 at a time, which is 6 P4.

Thus, the required number of words that can be formed using 4 letters at a time is

(ii) Number of words that can be formed by using all the letters of the word MONDAY at a time is equal to the number of permutations of 6 different objects taken 6 at a time is 6 P6 = 6! .

Thus, the required number of words that can be formed when all letters are used at a time = 6! = 6 × 5 × 4 × 3 × 2 ×1 = 720

(iii) In the given word, there are 2 different vowels, which have to occupy the rightmost place of the words formed.

This can be done only in 2 ways.

Since the letters cannot be repeated and the rightmost place is already occupied with a letter (which is a vowel), the remaining five places are to be filled by the remaining 5 letters.

This can be done in 5! ways.

Thus, in this case, required number of words that can be formed is 5! × 2 = 120 × 2 = 240

10: In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

In the word MISSISSIPPI, I appears 4 times, S appears 4 times, P appears 2 times, and M appears only once.

Therefore, number of distinct permutations of the letters in the given word

11! /4!4!2! = 11x10x9x8x7x6x5x4! /4!x4x3x2x1x2x1

= 11x10x9x8x7x6x5 /4x3x2x1x2x1 = 34650

There are 4 Is in the given word.

When they occur together, they are treated as a single object for the time being.

This single object together with the remaining 7 objects will account for 8 objects.

These 8 objects in which there are 4 Ss and 2 Ps can be arranged in 8! /4!2! ways i.e., 840 ways.

Number of arrangements where all Is occur together = 840

Thus, number of distinct permutations of the letters in MISSISSIPPI in which four Is do not come together = 34650 – 840 = 33810

11.In how many ways can the letters of the word PERMUTATIONS be arranged if the

(ii)vowels are all together,

(iii) there are always 4 letters between P and S?

In the word PERMUTATIONS, there are 2 Ts and all the other letters appear only once.

(i) If P and S are fixed at the extreme ends (P at the left end and S at the right end), then 10 letters are left.

Hence, in this case, required number of arrangements 10! /2! = 1814400

(ii) There are 5 vowels in the given word, each appearing only once. Since they have to always occur together, they are treated as a single object for the time being.

This single object together with the remaining 7 objects will account for 8 objects.

These 8 objects in which there are 2 Ts can be arranged in 8! /2!.

Corresponding to each of these arrangements, the 5 different vowels can be arranged in 5! ways.

Therefore, by multiplication principle, required number of arrangements in this case 8! /2! x5! = 2419200

(iii) The letters have to be arranged in such a way that there are always 4 letters between P and S.

Therefore, in a way, the places of P and S are fixed. The remaining 10 letters in which there are 2 T’s can be arranged in 10! /2! ways

Also the letters P and S can be placed such that there are 4letters between them in 2 x 7 = 14 ways

therefore by multiplication principle the required number of arrangements in this case = 10! /2! x 14 = 25401600

## Permutations and the Factorial Notation

The product of all the positive integers less than or equal to a number, say, is denoted by and is read “ factorial”. That is,

So as simple examples we have,

By definition, we have that

The factorial function , has domain equal to the non-negative integers. (Although in higher undergraduate mathematics there turns out to be a way to consider the factorial function for negative integers and fractional values). This notation is particularly handy when considering arrangements in lines or circles. Consider the example below.

### Example 3

In how many ways may 6 girls sit in a row?

### Solution 3

In this case we may go through a similar argument to that in example 2, and find out that the answer is equal to

In general when arranging distinct objects in a row (repetitions not allowed) we have that the number of arrangements or permutations of those objects are given by . Now, if we were to have objects, not all distinct, then this is a different matter, and in fact there does exist a formula for such a case. The formula is given below.

If there are objects, with objects being non-distinct and of a certain type and objects being non-distinct and of another type and objects being non-distinct and of another type and so on, we have that the number of arrangements of such objects in a row is given by,

Although this may not be immediately obvious, the reason for the division by the values of and is simply to eliminate all the rearrangements of the similar objects between themselves. Consider the example below that illustrates the use of this formula.

### Example 4

In how many ways may the letters of the word MAMMAL be rearranged?

### Solution 4

In this example, we note that there are M’s and that there are A’s, and that in total we have letters. Using the above formula we have that the number of arrangements of the letters is given by,

Suppose now that we wish to arrange objects in a circle. The difference between this and the above case is the lack of a defined starting point and a defined ending point. Because of this we must assign one of the objects being arranged, to be our reference point and it is then that we may arrange the remaining objects about the object chosen to be the reference point. Hence upon removing an object (to place as a reference point) from the total objects to be arranged, we end up with one less object and then these objects must be arranged in the same way as was considered in example 2. Hence we have that

If there are objects, all distinct, then the total number of ways in which these objects may be arranged about a circle is given by .

### Example 5

In how many ways may 5 married couples be arranged about a circular table?

### Solution 5

Since there are 5 married couples, it follows that there are people to arrange. Thus the total number of arrangements is given by,

Suppose now that we have objects among those being similar and another being similar and so on. The number of arrangements about a circle is now given by the total number of arrangements divided by the arrangements of the similar objects themselves. Hence we have that

If there are objects with similar and another similar and another similar etc. it follows that the total number of arrangements of those objects in a circle is given by

We now consider the arrangements of objects on a necklace, or a keychain. Suppose that we such to arrange distinct beads on a necklace. The total number of arrangements is given by the total number of unrestricted arrangements in a circle divided by . The division by is due to the symmetry of the necklace by being able to turn the necklace around and observe the exact same permutation. Hence we have that

Given objects, each distinct, the number of arrangements of these objects on a necklace is given by,

We have so far considered arrangements in rows and circles, given similarity between some of the objects. We shall now consider the number of arrangements of objects given certain conditions. Recall that a permutation of an object is simply an ordered selection or an arrangement. We thus have that,

The number of permutations of objects choosing at a time is given by

Notice that for the formula becomes,

Thus, if one is choosing all the objects in the set then the total number of permutations is simply the same as arranging all the objects into a line which should at this point be quite obvious. Also, this formula assumes that no repetitions are allowed. Let us consider an example to illustrate this formula’s use.

### Example 6

Find the number of letter arrangements of the letters of the word COMPLEX.

### Solution 6

Since every letter is distinct, it thus follows that the total number of such arrangements is given by

We are now going to consider some examples of questions that involve permutations given certain conditions. Students tend to have difficulty in grasping the techniques involved, since every question must be tackled on its own merits. The best way to master such questions is to expose oneself to as many examples as possible.

### Example 7

In horse racing, a trifecta is name given to selecting the first three greyhounds in a race in the correct order. In how many ways can this be done if there are 8 horses?

### Solution 7

In this question we are choosing a group of with order important from a set of . Using the formula for permutations gives,

### Example 8

In how many ways can 5 girls and 3 boys be arranged in a row if:
a) the boys must sit next to each other?

b) the boys must not sit next to each other?

### Solution 8

a) If the boys must sit next to each other, then we may treat the boys as one unit. Thus using this technique we have a total of objects to permute in a line which has a total number of permutations equal to . Now, the boys themselves may be arranged about themselves within their unit, and this can be done in ways. Thus by the fundamental counting principle, the total number of ways to arrange them in this fashion is,

b) Now, if the boys are NOT to sit next to each other, then we can simply consider the complementary event.

Let represent the boys sitting next to each other, and represent the boys not sitting next to each other.
Let represent the number of ways may occur.

Where is the complementary event. Thus we have that

Note: Do not fall into the common trap of listing cases and then counting the possible arrangements in each case. This is futile and tedious. If possible always look to the complementary event first.

### Example 9

In how many ways can the letters of the word MEETING be arranged if vowels and consonants occupy alternate places?

### Solution 9

Since there are 3 vowels and 4 consonants, then it follows that all arrangements have the form

where C represents a consonant and V represents a vowel.

Now, the vowels themselves may be arranged in their own possible positions in a total of ways owing to the fact that there exists two A’s and thus in this case we need to divide amongst the arrangements of the A’s themselves. The consonants may be arranged amongst themselves in their allocated positions in a total of ways. Hence the total number of arrangements is given by

The next example involves repetitions.

### Example 10

How many different number plates for cars exist if each contains 3 consonants of the alphabet followed by three digits?

### Solution 10

There are six places to fill. In the first three places, we have that a consonant may occur, implying that one of 21 possible letters may occur in each of the first three positions. In the last three positions one 10 possible numbers may go into each one of the positions. Note that in this case repetitions are allowed.

We shall now consider a probability question involving permutations. Recall that the probability of an event is given by

where the sample space is the set of all possible outcomes.

### Example 11

Mr. and Mrs. Smith and guests sit around a circular dinner table. Find the probability that the two hosts are together.

### Solution 11

So, to solve this question we need to find the total number of ways that the people may be arranged, and this becomes the number of elements in the sample space. We then need to find the no. of ways the hosts may be together.

To find the number of ways the hosts may be together, we simply consider them as one unit at which point we have 7 objects left, and then consider the possible rearrangements around circular dinner table. Recall that the couple may also be arranged amongst themselves.

Thus we have that the probability that the hosts are together is given by,

## Formulas at Work

To see the formulas at work, let’s look at the initial example. The number of permutations of a set of three objects taken two at a time is given by P(3,2) = 3!/(3 - 2)! = 6/1 = 6. This matches exactly what we obtained by listing all of the permutations.

The number of combinations of a set of three objects taken two at a time is given by:

C(3,2) = 3!/[2!(3-2)!] = 6/2 = 3. Again, this lines up exactly with what we saw before.

The formulas definitely save time when we are asked to find the number of permutations of a larger set. For instance, how many permutations are there of a set of ten objects taken three at a time? It would take awhile to list all the permutations, but with the formulas, we see that there would be:

P(10,3) = 10!/(10-3)! = 10!/7! = 10 x 9 x 8 = 720 permutations.

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## Combination

### Basic concepts

Combination is selection of rr objects from a set of nn distinct objects. In this case, there is no importance attached to the order of selection. Each unique set of rr objects selected form one combination.

As in each such combination the rr objects selected can be ordered among themselves in r!r!unique ways, if we order all the combinations in this way we would get the permutation of rr out of nn distinct objects. Thus, number of combinations multiplied with r!r! gives us number of permutations.
So,
Number of combinations,nCr=Number of Permutationsr!nCr=Number of Permutationsr!
=n!r!(n−r)!=n!r!(n−r)!

### Question 1: In how many ways can you select 3 books out of 5 available books?

Solution:
The number of ways 3 books can be selected out of 5 books is,
5C3=5!3!(5−3)!=5!3!2!=105C3=5!3!(5−3)!=5!3!2!=10
﻿

### Questions 2: In how many ways 4 members can be selected out of 8 members to form a committee such that 1 member is always selected?

Solution:
If 1 member is always selected in the committee then the combination choice problem is changed to selecting (4−1)=3(4−1)=3 members out of (8−1)=7(8−1)=7 members. The required number of ways then,
7C3=7!3!(7−3)!=7!3!4!=357C3=7!3!(7−3)!=7!3!4!=35

### Question 3: In how many ways 4 members can be selected out of 8 members to form a committee such that 2 members are always excluded?

Solution:
If two members are always excluded, the number of members to choose from reduces to (8−2)=6(8−2)=6 and the required number of combinations is,
6C4=6!4!(6−4)!=6!4!2!=156C4=6!4!(6−4)!=6!4!2!=15

## Combinations, Ho!

Combinations are easy going. Order doesn’t matter. You can mix it up and it looks the same. Let’s say I’m a cheapskate and can’t afford separate Gold, Silver and Bronze medals. In fact, I can only afford empty tin cans.

How many ways can I give 3 tin cans to 8 people?

Well, in this case, the order we pick people doesn’t matter. If I give a can to Alice, Bob and then Charlie, it’s the same as giving to Charlie, Alice and then Bob. Either way, they’re equally disappointed.

This raises an interesting point — we’ve got some redundancies here. Alice Bob Charlie = Charlie Bob Alice. For a moment, let’s just figure out how many ways we can rearrange 3 people.

Well, we have 3 choices for the first person, 2 for the second, and only 1 for the last. So we have $3 * 2 * 1$ ways to re-arrange 3 people.

Wait a minute… this is looking a bit like a permutation! You tricked me!

Indeed I did. If you have N people and you want to know how many arrangements there are for all of them, it’s just N factorial or N!

So, if we have 3 tin cans to give away, there are 3! or 6 variations for every choice we pick. If we want to figure out how many combinations we have, we just create all the permutations and divide by all the redundancies. In our case, we get 336 permutations (from above), and we divide by the 6 redundancies for each permutation and get 336/6 = 56.

which means “Find all the ways to pick k people from n, and divide by the k! variants”. Writing this out, we get our combination formula, or the number of ways to combine k items from a set of n:

Sometimes C(n,k) is written as:

Ex 7.3 Class 11 Maths Question 1.
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Solution.
Total digits are 9. We have to form 3 digit numbers without repetition.
∴ The required 3 digit numbers = 9 P3

Ex 7.3 Class 11 Maths Question 2.
How many 4-digit numbers are there with no digit repeated?
Solution.
The 4-digit numbers are formed from digits 0 to 9. In four digit numbers 0 is not taken at thousand’s place, so thousand’s place can be filled in 9 different ways. After filling thousand’s place, 9 digits are left. The remaining three places can be filled in 9P3 ways.
So the required 4-digit numbers
= 9 x 9 P3
= 9 x 504 = 4536.

Ex 7.3 Class 11 Maths Question 3.
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Solution.
For 3-digit even numbers unit place can be filled by 2, 4, 6 i.e in 3 ways. Then the remaining two places can be filled in 5 P2 ways.
∴ The required 3-digit even numbers
= 3 x 5 P2
= 60

Ex 7.3 Class 11 Maths Question 4.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
Solution.
The 4-digit numbers can be formed from digits 1 to 5 in 5 P4ways.
∴ The required 4 digit numbers = 5 P4 = 120 For 4-digit even numbers unit place can be filled by 2,4, i.e., in 2 ways. Then the remaining three places can be filled in 4 P3 ways.
∴ The required 4-digit even numbers
= 2 x 4 P3 = 2 x 24 = 48

Ex 7.3 Class 11 Maths Question 5.
From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?
Solution.
From a committee of 8 persons, we can choose a chairman and a vice chairman

Ex 7.3 Class 11 Maths Question 6.
Find n if n-1 P3: n P4 = 1 : 9.
Solution.

Ex 7.3 Class 11 Maths Question 7.
Find r if
(i) 5 Pr = 2 6 Pr-1
(ii) 5 Pr = 6 Pr-1
Solution.

Ex 7.3 Class 11 Maths Question 8.
How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
Solution.
No. of letters in the word EQUATION = 8
∴ No. of words that can be formed
= 8 P8 = 8!
=40320

Ex 7.3 Class 11 Maths Question 9.
How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if
(i) 4 letters are used at a time,
(ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?
Solution.
No. of letters in the word MONDAY = 6
(i) When 4 letters are used at a time.
Then, the required number of words
= 6 P4

(ii) When all letters are used at a time. Then the required number of words
= 6 P6 = 6!
= 720

(iii) All letters are used but first letter is a vowel.
So the first letter can be either A or O.
So there are 2 ways to fill the first letter & remaining places can be filled in 5 P5 ways.
∴ The required number of words
= 2 x 5 P5
= 2 x 5! =240.

Ex 7.3 Class 11 Maths Question 10.
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
Solution.
There are 11 letters, of which I appears 4 times, S appears 4 times, P appears 2 times & M appears 1 time.
∴ The required number of arrangements

= 10 x 10 x 9 x 7 x 5 = 34650 … (i)
When four I’s come together, we treat them as a single object. This single object with 7 remaining objects will account for 8 objects. These 8 objects in which there are 4S’s & 2P’s
can be rearranged in ways i.e. in 840 ways … (ii)
Number of arrangements when four I’s do not come together = 34650 – 840 = 33810.

Ex 7.3 Class 11 Maths Question 11.
In how many ways can the letters of the word PERMUTATIONS be arranged if the
(ii) vowels are all together,
(iii) there are always 4 letters between P and S?
Solution.
There are 12 letters of which T appears 2 times
(i) When words start with P and end with S, then there are 10 letters to be arranged of which T appears 2 times.
∴ The required words =

(ii) When vowels are taken together i.e E U A I O we treat them as a single object. This single object with remaining 7 objects will account for 8 objects, in which there w are 2Ts, which can be rearranged in ways. Corresponding to each of these arrangements the 5 vowels E, U, A, I, O can be rearranged in 5! = 120 ways. Therefore, by multiplication principle, the required number of arrangements = 20160 x 120 = 2419200.

(iii) When there are always 4 letters between P & S
∴ P & S can be at
1 st & 6 th place
2 nd & 7 th place
3 rd & 8 th place
4 th & 9 th place
5 th & 10 th place
6 th & 11 th place
7 th & 12 th place.
So, P & S will be placed in 7 ways & can be arranged in 7 x 2! = 14
The remaining 10 letters with 2T’s, can be arranged in ways.
∴ The required number of arrangements = 14 x 1814400= 25401600.

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## CBSE Mathematics

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### Permutation and Combination Class XI Chapter 7

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If an event can occur in m different ways, following which another event can occur in n different ways, following which another event can occur in p different ways, and so on. Then the total number of occurrence of the events in the given order is m x n x p…………..

Permutations when r epetition is allowed:

The number of permutations of n different objects taken all at a time, when repetition of objects is allowed is n n

Q ) How many positive numbers greater than 6000 and less than 7000 which are divisible by 5 if no digit is repeated. [Ans 112]