A permutation is an arrangement of all or part of a number of things in a definite order. For example, the permutations of the three letters a, b, c taken all at a time are abc, acb, bac, bca, cba, cab. The permutations of the three letters a, b, c taken two at a time are ab, ac, ba, bc, ca, cb.

Number of Permutations of n different things taken r at a time:

If n and r are positive integers such that 1≤r≤n, then the number of all permutations of n distinct things, taken r at a time is denoted by the symbol P (n, r) or .

when r = n,

Thus denotes the numbers of permutations of 8 different things taken 3 at a time, and^{ } denotes the number of permutations of 5 different things taken 3 at a time.

In permutations, the order of arrangement of elements is taken into account; when the order is changed, a different permutation is obtained.

Example: Find the value of and .

Solution:

Example: Find the total number of ways in which 4 persons can take their places in a cab having 6 seats.

Solution: The number of ways in which 4 persons can take their places in a cab having 6 seats: = ways.

## Permutation Derivations

**1. The total number of permutations of n different things taken all at a time is n!. **

Example: In how many ways 6 people can stand in a queue?

Solution: The number of ways in which 6 people can stand in a queue:

.

**2. The total number of arrangements of n different things taken r at a time, in which a particular things always occurs = **

Example: How many 4 digits number (repetition is not allowed) can be made by using digits 1-7 if 4 will always be there in the number?

Solution: Total digits (n) = 7

Now, as 4 will always occur, so remaining 3 digits could be selected from remaining 6 numbers in .

As in this arrangement, 4 will always occur, so 4 could be placed at any of the 4 places of 4 digit number. So total numbers of ways are 4 x 120 = 480.

This is same as above formula and could also be derived from the above formula as well. Total ways of making the number if 4 is always there =

= 480

**3. The total number of permutations of n different things taken r at a time in which a particular thing never occurs = **.

Example: How many different 3 letters words can be made by 5 vowels, if vowel ‘A’ will never be included?

Solution: Total letters (n) = 5

So total number of ways =

**4. The total number of permutations of n dissimilar things taken r at a time with repetitions = .**

Example: How many 3 digits number can be made by using digits 1 to 7 if repetition is allowed?

Solution: Total digits (n) = 7

So total ways = .

**The number of permutations of n things taken all at a time when p of them are alike and of one kind, q of them are alike and of second kind, all other being different, is:**

**The above theorem can be extended if in addition to the above r things are alike and of third kind and so on, then**

**Total permutation = **

Example: Find the number of permutations of the letters of the word ASSASSINATION.

Solution: We have 13 letters in all of which 3 are A’s, 4 are S’s, 2 are I’s and 2 are N’s

Therefore the number of arrangements =

Example: Find the number of permutations of the letters of the words ‘DADDY DID A DEADLY DEED’.

Solution: We have 19 letters in all of which 9 are Ds; 3 are As; 2 are Ys; 3 are Es and the rest are all distinct.

Therefore, the number of arrangements =

Example: How many different words can be formed with the letters of word ‘ORDINATE’?

- So that the vowels occupy odd places.
- Beginning with ‘O’.
- Beginning with ‘O’ and ending with ‘E’.

Solution:

1. ORDINATE contains 8 letters: 4 odd places, 4 vowels.

Number of arrangements of the vowels 4!.

Number of arranging consonants is 4!.

Number of words = 4! x 4! = (4*3*2*1)^2 = 576

2. When O is fixed we have only seven letters at our disposal.

Number of words = 7!=5040

3. When we have only six letters at our disposal, leaving ‘O’ and ‘E’ which are fixed. Number of permutations = 6!=720.

## Circular Permutation

So far we have discussed permutation of objects (or things) in a row. This type of permutations is generally known as linear permutations. If we arrange the objects along a closed curve viz. a circle, the permutations are known as circular permutations. As we have seen in the earlier sections of this chapter that every linear arrangement has a beginning and an end, but there is nothing like a beginning or an end in a circular permutation. Thus, in a circular permutation, we consider one object as fixed and the remaining objects are arranged as in case of linear arrangements.

**The number of ways of arranging n distinct objects around a circle is (n -1)! & the total number when taken r at a time will be .**

In circular permutation, anti-clockwise and clockwise order or arrangements are considered as distinct permutations.

Example: In how many ways can 7 Australians sit down at a round table?

Solution: 7 Australians can take their seat at the round table in (7-1)! = 6! ways.

Example: In how many ways can 12 persons among whom two are brothers be arranged along a circle so that there is exactly one person between the two brothers?

Solution: One person between the two brothers can be chosen in ways.

The remaining 9 persons can be arranged in .

The two brothers can be arranged in 2! ways.

Therefore, the total number of ways = (9!) (10)( 2!) = (10!)(2!).

**If there be no difference between clockwise and anticlockwise arrangements, the total no. of circular permutations of n things taken all at a time is and the total number when taken r at a time will be .**

Example: In how many ways can be garland of 10 different flowers be made?

Solution: Since, there is no difference between clockwise and anticlockwise arrangements, the total no. of ways

## Exercise

1. The total number of ways of answering 5 objective type questions, each question having 4 choices is:

- 256
- 512
- 1024
- 4096

2. The expression would be represented by

- P(n, n)
- P(n,2)
- P(n-2,2)
- P(n-2,0)

3. Which of the below expression would be equal to P(n,0)

- P(n, 1)
- P(n, n/2)
- n
- 1

4. A necklace is a circular string with several beads on it. It is allowed to rotate a necklace but not to turn it over. How many different necklaces can be made using 13 different beads?

- 13!
- 13!/2
- 12!
- 12!/2

5. Which of the below expression would be equal to P (n, r)

- C(n, 0)
- r! C(n, n-r)
- P(n, n-r)
- P(n, n/2)

6. P(n, n) equals

- n!
- 1
- 2^(n-1)

7. The number of ways of distributing 10 different books among 4 students (S1, S2, S3, S4) such that S1 and S2 get 2 books each and S3 and S4 get 3 books each is:

- 12600
- 15200

8. equals

- P(n, r)
- P(n-1,r+1)
- P(n+1,r)
- P(n+1,r+1)

9. P(5,r)=5, r equals

- 1
- 2
- 3
- 4

10. How many signals can be made with 5 different flags by raising them any number at a time?

- 375
- 475
- 275
- 325

11. P(r,2)=56, r equals

- 6
- 7
- 8
- 9

12. P(7,3)+ 3P(7,2) equals

- P(7,4)
- P(8,3)
- P(9,3)
- P(8,4)

13. Find the number of words of 5 letters such that each can be formed with the letters of the word “CHROMATE” if each letter may be repeated in any arrangement.

- 262144
- 4096
- 1024
- 32768

14. P(11,5)+ 5P(11,4) equals

- P(12,6)
- P(12,5)
- P(11,6)
- P(10,5)

15. P(13,3) + 3P(13,2) = P(14,r) then r equals

- 1
- 2
- 3
- 4

Rubavathi Manoharan says

Answers please. ….