Q45 / 160 · TS EAMCET 2013 · Mathematics
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
- A$11 ! 10 !$
- B$\frac{11 !}{6 ! 5 !}$
- C$\frac{10 ! 9 !}{5 !}$
- Correct answerD$\frac{11 ! 10 !}{5 !}$
Solution
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
$$
\begin{aligned}
S & =10 ! \times{ }^{11} P_6 \\
& =\frac{10 ! 11 !}{5 !}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.