Chapter 9.6, Problem 43E

To determine

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

Answer to Problem 43E

  $420$

Explanation of Solution

Given information:

Given group of letters

A, A, G, E, E, E, M

Calculation:

Let us consider the following group of letters

A, A, G, E, E, E, M

If a set of $n$ elements has $n_1$ of one kind of object, $n_2$ of 2^nd object…and so on with

  $n=n_1+n_2+\mathrm{...}+n_k$

Then number of distinguishable permutations of the $n$ object is given by

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

The above sequence of letters is made of $2A\text{'}s,3sE\text{'}s,1G,1M$ .

So, the distinguishable permutations are given by

  $\begin{array}{c}\frac{7!}{2!3!1!1!}=\frac{7!}{2\times 6}\\ =\frac{7\times 6\times 5\times 4\times 3\times 2\times 1}{2\times 6}\\ =7\times 5\times 4\times 3\\ =420\end{array}$