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

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

Permutations and Combinations

Solution:

There are $5$ vowels in the given word, which are $4 E's$ and $11$. Since, they have to always occur together, we treat them as a single object image for the time being. This single object together with $7$ remaining objects will account for $8$ objects. These $8$ objects, in which there are $3 N s$ and $2\,\,\, Ds$, 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 $= \frac{8!}{3! 2!} \times \frac{5!}{4!} = 16800$