Permutations on words

Question

How many six-letter words can be formed by using the letters of the word ‘PRESSES’?

So my doubt comes to which proceedure is valid to solve this problem:

• The book solution is: We omit in turn each of the four letters ‘P’, ‘R’, ‘E’ and ‘S’. This leaves six letters which we must then arrange in order. 1 If an S is omitted, there are then 2 Es and 2 Ss, so number of words = 6!, 2! × 2! = 180 2 If an E is omitted, there are then 3 Ss, so number of words = 6!, 3! = 120. 3 If P or R is omitted (2 cases), there are then 2 Es and 3 Ss, so number of words = 6!, 3! × 2! × 2 = 120 Hence the total number of words is $180 + 120 + 120 = 420$
• While my logic works like $\frac{7P6}{3!\cdot2}$ , 7Permute6 because it involves permuting 7 letters into 6 spaces and there are 3 letters repeated (the S's) and 2 letters repeated (the E's). And this gives you 420 too.

Is my logic/approach valid? Or is it just a coincidence that both methods get to the same answer, but mine is ilogical?

Answer 1

Your logic gives the correct answer only because exactly one letter is omitted. The result is the same as the number of seven-letter words, $\frac{7!}{3!2!}$. There is a bijection between six- and seven-letter words: PRESSE <=> PRESSES