How many ways are there to arrange 10 distinct men and 10 distinct women on 7 seats, such that no 2 women are together? The part where the number of seats are limited is causing a difficulty. Seating all of the men and women with this restriction is easy. But the restriction on number of seats mixed it all up.
 A: The condition that no two women sit together warrants that you pick m number of men out of 10 men and n women out of 10 women such that n = 1, 2 or 3 or 4.
For n = 1:  The number of ways you can choose 1 of 10 women is ${10\choose1}$ and the rest of the six men are chosen in ${10\choose6}$.  Having chosen this, now we need to make sure that these women don't sit together. For n = 1, it is straight 7 ways you can arrange the woman and 6! ways you can arrange the men in remainder of the six seats.  Thus the total arrangement is ${10\choose1}{10\choose6}7.6!$
For n = 2: In a similar argument.The number of ways you can choose 2 of 10 women is ${10\choose2}$ and the rest of the five men are chosen in ${10\choose5}$.  Having chosen this, now we need to make sure that these women don't sit together.we want to fill $2$ women around $5$ men, then we have a total number of arrangements that is  ${(5+1)\choose 2}. 2! 5!$.  Thus the total arrangement is ${10\choose2}{10\choose5}{6\choose2}2!.5!$
For n= 3:In a similar argument.The number of ways you can choose 3 of 10 women is ${10\choose3}$ and the rest of the 4 men are chosen in ${10\choose4}$.  Having chosen this, now we need to make sure that these women don't sit together.we want to fill $3$ women around $4$ men, then we have a total number of arrangements that is  ${(4+1)\choose 3}. 3! 4!$.  Thus the total arrangement is ${10\choose3}{10\choose4}{5\choose3}3!.4!$
For n = 4: In a similar argument.The number of ways you can choose 4 of 10 women is ${10\choose4}$ and the rest of the 3 men are chosen in ${10\choose3}$.we want to fill $4$ women around $3$ men, then we have a total number of arrangements that is  ${(3+1)\choose 4}. 4! 3!$.  Thus the total arrangement is ${10\choose4}{10\choose3}{4\choose4}4!.3!$
The final answer is sum up all cases.
