Chapter 9, Problem 15RE

Solution

To calculate: The number of code words of any length that can be spelled out using tiles of five different letters (including single-letter code words).

The resultant value is 325 codes.

Given information:

There are five tiles of different letters that has to be used to spell out a code of any length (including single-letter code words).

Formula used: The number of permutations of n objects taken r at a time denoted by $P_{n_r}$ and given by $P_{n_r}=\frac{n!}{\left(n-r\right)!}$ .

Calculation:

There are five tiles of different letters. If these are to be used to spell a code of any length then the outcomes can be: One letter, two letters, three letters, four letters or five letters

The number of different words spelled from 5 different letters is: $P_{5_1}+P_{5_2}+P_{5_3}+P_{5_4}+P_{5_5}$

Rewrite the expression using the formula $P_{n_r}=\frac{n!}{\left(n-r\right)!}$ ,

  $\begin{array}{c}{P}_{5_1}+P_{5_2}+P_{5_3}+P_{5_4}+P_{5_5}=\frac{5!}{\left(5-1\right)!}+\frac{5!}{\left(5-2\right)!}+\frac{5!}{\left(5-3\right)!}+\frac{5!}{\left(5-4\right)!}+\frac{5!}{\left(5-5\right)!}\\ =\frac{5\cdot 4!}{4!}+\frac{5\cdot 4\cdot 3!}{3!}+\frac{5\cdot 4\cdot 3\cdot 2!}{2!}+\frac{5\cdot 4\cdot 3\cdot 2\cdot 1!}{1!}+\frac{5\cdot 4\cdot 3\cdot 2\cdot 1!}{0!}\\ =5+20+60+120+120\\ =325\end{array}$

Therefore, 325 codes can be spelled out from 5 different letters tiles.