Permutation and Combination with restrictions

Question

There are 7 chairs. 4 are reserved for men and 2 are reserved for ladies. Remaining 1 can be occupied by anybody - either a lady or a man. In how many ways, can 6 men and 3 ladies be seated on these 7 chairs. Thanks.

Answer 1

Hint: You can seat either four men or five. For permutations involving four, how many ways are there to seat four men out of six in the four chairs? Then you seat three women in the other three chairs. Do the same for five.

Answer 2

First choose the 4 men to sit in the reserved chairs $\binom 64$, multiply by $4!$ since they can sit in any order in those chairs. Then choose the two women to sit in the reserved chairs: $\binom 32$ and multiply by $2!$, since the two can sit in any order in the reserved chairs. Then choose 1 of the remaining 3 persons (one of the remaining two men and one woman) $\binom 31$. There is only one unreserved sit to sit him/her.

In all, that gives us $$\binom 64\cdot 4! \cdot \binom 32 \cdot 2! \cdot \binom 31= 15 \cdot 24 \cdot 3 \cdot 2\cdot 3 = 6480\; \text{ ways to sit them as required.}$$