Chapter 13.2, Problem 52AYU

To determine

To find: How many different 11-letter words (real or imaginary) can be formed from the letters in the word MATHEMATICS?

Answer to Problem 52AYU

4989600

Explanation of Solution

Given:

The word MATHEMATICS.

Formula used:

$C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

Assuming that the words can be repeated,

This is a permutation with repetition.

Each word formed will have 11 letters: 2 M’s, 2 A’s, 2 T’s, 1 H, 1 E, 1 I, 1C and 1 S.

To construct each word, we need to fill in 11 positions with the 8 letters:

The process of forming a word consists of seven tasks.

Choose the positions for the $2M=C\left(11,2\right)$.

Choose the positions for the $2A’s= C\left(9,2\right)$.

Choose the positions for the $2T’s=C\left(7,2\right)$.

Choose the position for the $1H=C\left(5,1\right)$.

Choose the position for the $1E = C\left(4,1\right)$.

Choose the position for the $1I = C\left(3,1\right)$.

Choose the position for the $1C = C\left(2,1\right)$.

Choose the position for the $1S = C\left(1,1\right)$.

Number of possible ways that can be formed is

$C\left(11,2\right)*C\left(9,1\right)* C\left(8,2\right)*C\left(6,2\right)*C\left(4,1\right)*C\left(3,1\right)*C\left(2,1\right) *C\left(1,1\right)$.

$\frac{11!}{9!*2!}*\frac{9!}{2!*7!}*\frac{7!}{2!5!}*\frac{5!}{4!}*\frac{4!}{3!}*\frac{3!}{2!}*2!=\frac{11!}{2!*2!*2!}=4989600$.