(D) First,we arrange $10$ men in a row. The number of ways to arrange $10$ men is $10!$. There are $11$ gaps created by $10$ men (including the ends)

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(D) First,we arrange $10$ men in a row. The number of ways to arrange $10$ men is $10!$. There are $11$ gaps created by $10$ men (including the ends) where $6$ women can be seated so that no two women sit together. The number of ways to choose $6$ gaps out of $11$ and arrange $6$ women is given by $^{11}P_6$. Total number of ways $= 10! \times ^{11}P_6 = 10! \times \frac{11!}{(11-6)!} = \frac{10! 11!}{5!}$.

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

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$10$ 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
$11! 10!$
B
$\frac{11!}{6! 5!}$
C
$\frac{10! 9!}{5!}$
D
$\frac{11! 10!}{5!}$

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

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$10$ 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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