In how many ways 5 men and 4 women can be arranged in a line so that

Solution provided by sandeep800 is correct.

Consider this:
5 men can be arranged in 5! number of ways;

Consider one particular arrangement of these men and empty slots as follows: *m*m*m*m*m*. Now, if we place 4 women in either of 4 slots out of 6 then no two women will be together --> \(C^4_6\) is the # of ways to choose in which 4 slots out of 6 these women will be placed and 4! is # of arrangements of them in these slots.

So total # of arrangements is: \(5!*C^4_6*4!\), which is the same as \(5!*P^4_6\).

As for 6p4: \(P^4_6=\frac{6!}{(6-4)!}=360\). # of ways to choose k distinct object out of n distinct object when the order of the chosen objects matters is given by the permutation formula: \(P^k_n=\frac{n!}{(n-k)!}\).

Hope it's clear.