Permutations

                                    Permutations

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, bca, bac, 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 ⁿP_r

ⁿP_r  = n!/(n-r)! = n(n-1) (n-2)…….(n-r+1)

When r=n,  ⁿP_r   = ⁿP_n   =n(n-1)(n-2) ... 1=n!.

Thus ⁸P₃ denotes the numbers of permutations of 8 different things taken 3 at a time, and ⁵P₅ denotes the number of permutations of 5 different things taken 5 at a time.

Important

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

E10. Find the values of ⁵P₁, ⁵P₂, ⁵P₃, ⁵P₄, ⁵P₅ and ¹⁰P₇

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

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

=⁶p₄=6x5x4x3=360ways.

Some Important Formulae

Ø  The total number of permutations of n different things taken all at a time is n!

E12.In how many ways 6 people can stand in a queue?

        Sol. The number of ways in which 6 people can stand in a queue = 6! = 720.

Ø  The total number of arrangements of n different things taken r at a time, in which a particular thing always occurs      =       r X ^n-1P_r-1

E13. 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?

Sol. Total digits (n) = 7

Total ways of making the number if 4 is always there =   r x ^n-1P_r-1  = 4 x ⁶P₃ = 480.

Ø  The total number of permutations of n different things taken r at a time in which a particular thing never occurs    = r x ^n-1P_r

E14. How many different 3 letter words can be made by 5 vowels, if vowel 'A' will never be included?

Sol. Total letters (n) = 5

So total number of ways = ^n-1P_r = ^5-1P₃ = ⁴P₃ = 24.

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

E15.                                                                             How many 3 digits number can be made by using digits 1 to 7 if repetition is allowed?

    Sol. Total digits (n) = 7

So total ways = 7³ = 343.

Ø 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      n!/P! q!

Remark: 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 =  n!/P! q! r!......

E16.  Find the number of permutations of the letters of the word ASSASSINATION.

   Sol. 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

    The number of arrangements  = 13!/3! 4! 2! 2!

E17.  Find the number of permutations of the letters of the words 'DADDY DID A DEADLY DEED'.

     Sol. 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     = 19!/9!3!2!3!

E18. How many different words can be formed with the letters of word 'ORDINATE'?

(1)    so that the vowels occupy odd places.

(2)    beginning with'0'.

(3)    beginning with 'O' and ending with'E'.

     Sol. (1) ORDINATE contains 8 letters:

4 odd places,  4 vowels.

=number of arrangements of the vowels 4!

Also number of arranging consonants is 4!

Number of words = 4! x 4! =(4x3x2x1)²=576.

 (2) When 0 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 '0' 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 are 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  ⁿP_r/r

Note: In the above, anti-clockwise and cIockwise  order of arrangements are considered as distinct permutations.

E19.In how many ways can 7 Australians sit down at a round table?

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

E20.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?

Sol. One person between the two brothers can be chosen in ¹⁰ P₁ = 10 ways. The remaining 9 persons can be arrangedin ⁹P₉ = 9! . The two brothers can be arranged in 2! ways.

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

Important

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    (n-1)!/2   and the total number when taken r at a time will be      ⁿP_r/2r

E21.In how many ways can a garland of 10 different  flowers be made?

Sol. Since, there is no difference between clockwise and anticlockwise arrangements, the total no. of ways = (10 - 1)!/2 = 9!/2.