If all the letters of the word COMBINATION ar | vedclass

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Question

(C) The word $COMBINATION$ has $11$ letters: $C, O, M, B, I, N, A, T, I, O, N$. The frequency of letters is: $C:1, O:2, M:1, B:1, I:2, N:2, A:1, T:1$.

Vedclass · Accepted Answer

$2(8!)$

Vedclass · Answer

(C) The word $COMBINATION$ has $11$ letters: $C, O, M, B, I, N, A, T, I, O, N$. The frequency of letters is: $C:1, O:2, M:1, B:1, I:2, N:2, A:1, T:1$. Total vowels are $O, I, A, I, O$ ($5$ vowels) and consonants are $C, M, B, N, T, N$ ($6$ consonants).Step $1$: $C$ and $N$ occupy the end positions. There are $2$ ways to place $C$ and $N$ at the ends ($C...N$ or $N...C$).Step $2$: Remaining $9$ letters are $O, M, B, I, A, T, I, O, N$ (if $C$ and $N$ are fixed at ends). The middle position is the $6$th position. There are $9$ letters to arrange in the remaining $9$ spots.Step $3$: Total arrangements with $C, N$ at ends $= 2 	imes \frac{9!}{2!2!2!} = 2 	imes \frac{362880}{8} = 90720$.Step $4$: Arrangements where a vowel is in the middle: The middle position can be filled by $O, I, A, I, O$ ($5$ choices). If a vowel is fixed in the middle,we arrange the remaining $8$ letters. Total $= 2 	imes (5 	imes \frac{8!}{2!2!}) = 2 	imes 5 	imes 10080 = 100800$ (Wait,adjusting for repetitions: $2 	imes (2 	imes \frac{8!}{2!2!} + 1 	imes \frac{8!}{2!2!2!} + 2 	imes \frac{8!}{2!2!}) = 2 	imes (20160 + 5040 + 20160) = 90720$ is total. Subtracting cases with vowel in middle: $2 	imes (2 	imes \frac{8!}{2!2!} + 1 	imes \frac{8!}{2!2!2!}) = 2 	imes (20160 + 5040) = 50400$.Result $= 90720 - 50400 = 40320 = 2(8!)$.

If all the letters of the word $COMBINATION$ are arranged in all possible ways to form $11$ letter words (with or without meaning),then the number of words among them in which $C$ and $N$ occupy the end positions and no vowel appears exactly in the middle position is

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