How many permutations of the word $STRESSLESSNESS$ begin OR end with an $E$?

Question

How many permutations of the word $STRESSLESSNESS$ begin OR end with an $E$?

Correct me if I'm wrong, but you would have to subtract the permutations where $E$ begins AND ends the permutation?

Answer 1

• $E\underbrace{------------}_{12 \text {letter}}E$
  Result: $\frac{12}{7!}$ , since there are seven $S$'s among the remaining letters(STRSSLSSNESS) we divide by $7!$ to eliminate overcounting

• $E\underbrace{-------------}_{13 \text {letter}}$
  Result: $\frac{13}{2!7!}$ ,since there are seven $S$'s and two $E$'s among the remaining letters(STRSSLESSNESS) we divide by $7!2!$ to eliminate overcounting

• $\underbrace{-------------}_{13 \text {letter}}E$
  Result: $\frac{13}{2!7!}$ , again since there are seven $S$'s and two $E$'s among the remaining letters(STRSSLESSNESS) we divide by $7!2!$ to eliminate overcounting

But remember $2nd$ and $3rd$ cases contain $1st$ case as well. So, we need to handle that overcounting too by substracting $1st$ result from $2nd$ and $3rd$ results.

So, in total: $\frac{12}{7!}+(\frac{13}{2!7!}-\frac{12}{7!})+ (\frac{13}{2!7!}-\frac{12}{7!})$

Answer 2

STRESSLESSNESS is 14 letters long.

If we take away an e, it becomes 13 letters long.

The permutations of this 13 letter string is:

$$\frac{13!}{7!2!}$$

Since the E can occur at the beginning or the end, we multiply this by 2 to get: $$\frac{13!}{7!}$$

We then take into account a possibility of an E at the beginning and at the end:

$$\frac{13!}{7!}-\frac{12!}{7!}=1140480$$

Answer 3

One way of doing it is this:

number that begins with an E, plus number than ends with an E, minus number that both begins and ends with an E.

Another way is this:

First find the total number of permutations, then subtract the number that begin and end with a non-E.