Question

The letters of the word "INDEPENDENCE" are arranged in all possible ways.

The number of words in which the 'D's do not come together is

• A. $\frac{10!}{4!3!}$
• B. $\frac{10!}{4!3!}\frac{{}^{10}{P}_2}{2!}$
• C. $\frac{10!}{4!3!}{.}^{10}{P}_2$
• D. $\frac{10!}{4!3!}\frac{{}^{11}{P}_2}{2!}$

Answer 1

The correct option is D $\frac{10!}{4!3!}\frac{{}^{11}{P}_2}{2!}$

The word INDEPENDENCE has 4E,2D,3N and I,P,C
that is $EEEEDDNNNIPC$
now let's permute all the words other than D.This can be done in $\frac{10!}{4!\times 3!}$ ways.
now the placing of these 10 words creates 11 gap.So any 2 of these gap can be chosen to fill the Ds.This can be done in $\left(\begin{array}{c}11\\ 2\end{array}\right)$ ways.
Hence no. of words in which Ds don't come together are $\frac{10!}{4!\times 3!}\times \left(\begin{array}{c}11\\ 2\end{array}\right)$
Hence, option 'D' is correct.