Question

6 women and 5 men are to be seated in a row so that no 2 men can sit together. Number of ways they can be seated is

• A. 5!$\times {}^7{P}_5$
• B. 6!$\times {}^6{P}_5$
• C. 5!$\times {}^7{P}_5$
• D. 6!$\times {}^7{P}_5$

Answer 1

The correct option is D 6!$\times {}^7{P}_5$

Condition is that no 2 men can sit together.

One of the ways to achieve this is to first make 6 women to sit together and arrange 5 men in the gaps as shown below.

6 women can be made to sit together in 6! ways.

_W_W_W_W_W_W_

Observe that there are 7 places and 5 men are to be seated in these 7 places.

In other words, for 5 men, we have to arrange 5 places from the available 7 places. This can be done in ${}^7{P}_5$ ways

So, the total number of ways they can be seated = 6! $\times {}^7{P}_5$