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?
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