The number of ways of arranging all the lette | vedclass

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Question

(A) The word "$COMBINATIONS$" has $12$ letters: $C, O, M, B, I, N, A, T, I, O, N, S$.Consonants: $C, M, B, N, N, T, S$ ($7$ letters,with $N$ repeating

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$\frac{7!6!}{(2!)^4}$

Vedclass · Answer

(A) The word "$COMBINATIONS$" has $12$ letters: $C, O, M, B, I, N, A, T, I, O, N, S$.Consonants: $C, M, B, N, N, T, S$ ($7$ letters,with $N$ repeating twice).Vowels: $O, I, A, I, O$ ($5$ letters,with $O$ repeating twice and $I$ repeating twice).First,arrange the $7$ consonants in a circle. The number of ways to arrange $n$ items in a circle is $(n-1)!$. Since $N$ repeats twice,the number of ways is $\frac{(7-1)!}{2!} = \frac{6!}{2!}$.There are $7$ gaps created between these $7$ consonants. We need to place the $5$ vowels in these $7$ gaps such that no two vowels are together. The number of ways to choose $5$ gaps out of $7$ is ${ }^{7}C_{5}$.The $5$ vowels can be arranged in these $5$ chosen gaps in $5!$ ways. Since $O$ and $I$ each repeat twice,the number of arrangements is $\frac{5!}{2!2!}$.Total number of ways $= \frac{6!}{2!} 	imes { }^{7}C_{5} 	imes \frac{5!}{2!2!} = \frac{6!}{2!} 	imes \frac{7!}{5!2!} 	imes \frac{5!}{2!2!} = \frac{7!6!}{(2!)^4}$.

The number of ways of arranging all the letters of the word "$COMBINATIONS$" around a circle so that no two vowels come together is

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