Question

In the word 'ENGINEERING' if all E's are not together and N's are always together, then number of permutations is

• A. $\frac{9!}{2!2!}-\frac{7!}{2!2!}$
• B. $\frac{9!}{3!2!}-\frac{7!}{2!2!}$
• C. $\frac{9!}{3!2!2!}-\frac{7!}{2!2!2!}$
• D. $\frac{9!}{3!2!2!}-\frac{7!}{2!2!}$

Answer 1

Answer: (D)

$\frac{9!}{3!2!2!}-\frac{7!}{2!2!}$

Lets take $NNN=$ One unit

So remaining $=8$ units. Therefore we have total 9 units

Thus number of words with all N's together and E's not together $=$ Total permutations $-$ E's come together.

Total permutation $=\frac{9!}{3!2!2!}$

Permutations when E's comes together $=\frac{7!}{2!2!}$