Find the number of ways in which all eight letters of the word NEEDLESS can be arranged if the three Letters E must placed together and the two letters S must not be placed together?
What I have done:
Finding all possible arrangements: $8! =40320$
S must not placed together (treating 2 S's as one): $7! 2!$
Finally, $8! – 7! 2! = 30240$
And placing 3 E's together: $6!$
Final Arrangements: $6!+ 8! -7! 2!$
But by this technique, I am not getting the answer. Please let me know what should I do?
Best Answer
Consider EEE as a single letter. So the number of arrangements of N,EEE, D,L,S,S without the restriction on the letter S is $\frac{6!}{2!}$. From this number we subtract the number of arrangements of N,EEE, D,L,SS where SS is a single letter, i.e. the two S stay together, that is $5!$. Hence the final result is $$\frac{6!}{2!}-5!=240.$$