# permutation

• Dec 31st 2008, 05:34 AM
permutation
Find the number of words formed by permuting all the letters of the word EXERCISES .
• Dec 31st 2008, 05:48 AM
Mush
Quote:

Find the number of words formed by permuting all the letters of the word EXERCISES .

$\displaystyle n ! = n \times (n-1) \times (n-2) \times (n-3) \times ... \times 1$ where $\displaystyle n$ is the number of letters in the word.

Assuming that by "words" you mean combinations of letters, rather than coherent words with literary meanings.
• Dec 31st 2008, 05:51 AM
Isomorphism
Quote:

Find the number of words formed by permuting all the letters of the word EXERCISES .

Do you know that the number of arrangements of objects where $\displaystyle a_1$ are of one kind,$\displaystyle a_2$ are of another kind,... $\displaystyle a_n$ are of the nth kind is $\displaystyle \frac{(a_1 + a_2 + ... a_n)!}{a_1 ! a_2 ! ... a_n!}$
• Dec 31st 2008, 06:23 AM
Soroban

This is a permutation with some duplicated objects.
There is a formula for this, but you can derive it yourself.

Quote:

Find the number of words formed by permuting all the letters of the word EXERCISES.

If we had 9 different letters, there would be $\displaystyle 9!$ possible words.

But there are 3 identical E's.
Switching them would not produce a new word.
So our answer is too large by a factor of $\displaystyle 3!$

. . (The number of ways 3 objects can be arranged.)

And there are 2 identical S's.
They can be switched in $\displaystyle 2!$ ways without producing a new word.
So our answer is also too large by a factor of $\displaystyle 2!$

So the answer is: .$\displaystyle \frac{9!}{3!\,2!} \:=\:30,\!240$