(C) First,we arrange $10$ men in a row. The number of ways to arrange $10$ men is $10!$. This creates $11$ possible gaps (including the ends) where th

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(C) First,we arrange $10$ men in a row. The number of ways to arrange $10$ men is $10!$. This creates $11$ possible gaps (including the ends) where the $6$ women can be seated to ensure no two women sit together. The number of ways to choose $6$ gaps out of $11$ and arrange the $6$ women is given by the permutation formula $^{11}P_6$. $^{11}P_6 = \frac{11!}{(11-6)!} = \frac{11!}{5!}$. Therefore,the total number of ways is $10! \times \frac{11!}{5!} = \frac{10! 11!}{5!}$.

Class 11 Mathematics Permutation and Combination Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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Ten men and $6$ women are to be seated in a row so that no two women sit together. The number of ways they can be seated is:

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A
$11! \times 10!$
B
$\frac{11!}{6! 5!}$
C
$\frac{10! 11!}{5!}$
D
$\frac{11! 10!}{6!}$

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Permutation and Combination

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Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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Ten men and $6$ women are to be seated in a row so that no two women sit together. The number of ways they can be seated is:

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