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?
On
As Robert pointed out, you are not correctly counting the arrangements with all the E's together.
But beyond that, your technique is overcounting. You are calculating: $$\text{# arrangements with E's together}+ \text{# arrangements with S's apart.}$$ First of all, this counts some arrangements that have the E's together but don't have the S's apart, which you don't want to count, and arrangements that have the S's apart but don't have the E's together, which you also don't want to count. The arrangements that you do want to count have both the E's together and the S's apart, and these are all counted twice. Instead you should count: $$\text{# arrangements with E's together} - \text{# arrangements with E's together and S's together.}$$ What this leaves is the number of arrangements that have the E's together and don't have the S's together, i.e., have the S's apart.
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 $$\frac{6!}{2!}-5!=240.$$