Chapter 8.5, Problem 50E

To determine

To find: the number of distinguishable permutations of the group of letters.

Answer to Problem 50E

The numbers of distinguishable permutations are $\frac{n!}{n_1!\cdot n_2!\cdot n_3!\cdot n_4}=34650$ .

Explanation of Solution

Given information:

Given group of letters

M,I,S,S,I,S,S,I,P,P,I

Calculation:

Let us consider the following group of letters

M, I, S, S, I, S, S, I, P, P, I

Find the number of distinguishable permutations of the above group of letters.

Use formulas of distinguishable permutations,

  $\frac{n!}{n_1!\cdot n_2!\cdot n_3!\cdot n_4}\mathrm{.......}(1)$

Suppose a set of $n$ object has $n_1$ of one kind of object, $n_2$ of a second kind of object.

Here,

  $\begin{array}{l}{n}_1=1\\ n_2=4\\ n_3=4\\ n_4=2\end{array}$

Substitute value in equation (1)

  $\begin{array}{c}\frac{n!}{n_1!\cdot n_2!\cdot n_3!\cdot n_4}=\frac{11!}{1!\times 4!\times 4!\times 2!}\\ =\frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4!}{4!\times 4\times 3\times 2\times 1\times 2\times 1}\\ =11\times 10\times 9\times 7\times 5\\ =34650\end{array}$

Hence, the numbers of distinguishable permutations are $\frac{n!}{n_1!\cdot n_2!\cdot n_3!\cdot n_4}=34650$ .