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Question: Answered & Verified by Expert
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
MathematicsPermutation CombinationAP EAMCETAP EAMCET 2013
Options:
  • A $11 ! 10 !$
  • B $\frac{11 !}{6 ! 5 !}$
  • C $\frac{10 ! 9 !}{5 !}$
  • D $\frac{11 ! 10 !}{5 !}$
Solution:
2675 Upvotes Verified Answer
The correct answer is: $\frac{11 ! 10 !}{5 !}$
W W W W W W W W W First, we arrange 10 men in a row at alternate position.
So, number of ways formula $=10$ !
Now, 6 women can arrange in 11 positions
So, number of ways for women $={ }^{11} P_6$
Required number of
$$
\begin{aligned}
S & =10 ! \times{ }^{11} P_6 \\
& =\frac{10 ! 11 !}{5 !}
\end{aligned}
$$

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