Chapter 8.5, Problem 86E

To determine

To find: the number of distinguishable permutation of the letters A, A, A, A, B, B and B.

Answer to Problem 86E

The number of distinguishable permutation of the letters A,A,A,A,B,B and B is 35.

Explanation of Solution

Given information: Given letters are A, A, A, A, B, B and B.

Formula used:

Suppose a set of n objects has $n_1$ one kind of object, $n_2$ of a second kind and so on with

  $n=n_1+n_2+n_3+\mathrm{..........}+n_k.$

The number of distinguishable permutation of the n object is given by,

  $\frac{n!}{n_1!\times n_2\times n_3!\mathrm{.............}{n}_k!}.$

Calculation:

Total number of letters =4 A letters +3 B letters=7 letters.

The number of distinguishable permutation of the letters A, A, A, A, B, B and B,

  $=\frac{7!}{4!\times 3!}=\frac{7\times 6\times 5\times 4!}{4!\times 3!}=\frac{7\times 6\times 5}{3\times 2\times 1}=35.$

So, the number of distinguishable permutation of the letters A, A, A, A, B, B and B is 35.