Question

Find the number of arrangements of the letters of the word INDEPENDENCE when all the vowels always occur together.

• A. $16550$
• B. $16800$
• C. $16850$
• D. $16500$

Answer 1

Answer: (B)

There are $5$ vowels in the given word, which are $4E^{\prime }s$ and $11$. Since, they have to always occur together, we treat them as a single object for the time being. This single object together with $7$ remaining objects will account for $8$ objects. These $8$ objects, in which there are $3Ns$ and $2Ds$, can be rearranged in $\frac{8!}{3!2!}$ ways. Corresponding to each of these arrangements, the $5$ vowels $E,E,E,E$ and $I$ can be rearranged in $\frac{5!}{4!}$ ways .Therefore, by multiplication principle, the required number of arrangements $=\frac{8!}{3!2!}\times \frac{5!}{4!}=16800$