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Permutation

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 ^n P _r.

^n P _r = \dfrac{n!}{(n-r)!} = n(n-1)(n-2) ... (n-r+1)

when r = n, ^n P _r= ^n P _n = n(n-1)(n-2) ... (1) = n!

Thus ^8 P _3 denotes the numbers of permutations of 8 different things taken 3 at a time, and ^5 P _3 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 ^5 P _1 , ^5 P _2 , ^5 P _3 , ^5 P _4 , ^5 P _5 and ^{10} P _7.

Solution:
^5 P _1 = 5
^5 P _2 = 5 \times 4 = 20
^5 P _3 = 5 \times 4 \times 3 = 60
^5 P _4 = 5 \times 4 \times 3 \times 2 = 120
^5 P _5 = 5 \times 4 \times 3 \times 2 \times 1 = 120
^10 P _7 = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 = 604800

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: = ^6 P _4 = 6 \times 5 \times 4 \times 3 = 360 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:

= 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720.

2. The total number of arrangements of n different things taken r at a time, in which a particular things always occurs = r \times ^{n-1} P _{r-1}

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 ^6 P _3 = 120 .

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 = r \times ^{n-1} P _{r-1}
= 4 \times ^{6} P _{3}
= 480

3. The total number of permutations of n different things taken r at a time in which a particular thing never occurs = ^{n-1} P _{r}.

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 = ^{n-1} P _{r} = ^{5-1} P _{3} = ^{4} P _{3} = 24

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

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 = 7^3 = 343 .

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:

\dfrac {n!}{p! \times q!}

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 = \dfrac {n!}{p! \times q! \times r! ... }

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 = \dfrac {13!}{3! \times 4! \times 2! \times 2! }

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 = \dfrac {19!}{9! \times 3! \times 2! \times 3! }

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

  1. So that the vowels occupy odd places.
  2. Beginning with ‘O’.
  3. 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 \dfrac{ ^n P _r }{r} .

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 ^{10} P _1 = 10 ways.

The remaining 9 persons can be arranged in ^9 P _9 = 9! .

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 \dfrac{(n-1)!}{2} and the total number when taken r at a time will be \dfrac{^n P _r}{2r} .

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 = \dfrac{(10-1)!}{2} = 9!/2

Exercise

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

  1. 256
  2. 512
  3. 1024
  4. 4096

2. The expression \dfrac{n!}{(n-2)!} would be represented by

  1. P(n, n)
  2. P(n,2)
  3. P(n-2,2)
  4. P(n-2,0)

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

  1. P(n, 1)
  2. P(n, n/2)
  3. n
  4. 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?

  1. 13!
  2. 13!/2
  3. 12!
  4. 12!/2

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

  1. C(n, 0)
  2. r! C(n, n-r)
  3. P(n, n-r)
  4. P(n, n/2)

6. P(n, n) equals

  1. 2^n
  2. n!
  3. 1
  4. 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:

  1. 12600
  2. 15200
  3. ^{10}P _4
  4. \dfrac{10!}{2!2!3!3!}

8. P(n-1,r) +rP(ni1,r-1) equals

  1. P(n, r)
  2. P(n-1,r+1)
  3. P(n+1,r)
  4. P(n+1,r+1)

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

  1. 1
  2. 2
  3. 3
  4. 4

 

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

  1. 375
  2. 475
  3. 275
  4. 325

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

  1. 6
  2. 7
  3. 8
  4. 9

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

  1. P(7,4)
  2. P(8,3)
  3. P(9,3)
  4. 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.

  1. 262144
  2. 4096
  3. 1024
  4. 32768

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

  1. P(12,6)
  2. P(12,5)
  3. P(11,6)
  4. P(10,5)

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

  1. 1
  2. 2
  3. 3
  4. 4


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