From among six couples, a committee of $5$ members is to be formed. If the selected committee has no couple, then in how many ways can the committee be formed?
My approach is to count things step by step like first compute the number of ways such that there is no male, then number of ways there is only one male and his partner is not, two males and their partners are not ... to number of ways there are no males on the committee.
Put in mathematics which looks like: $\binom{6}{0} \times \binom{6}{1}+\binom{6}{1} \times \binom{5}{4}+\binom{6}{2} \times \binom{4}{3}+ \cdots +\binom{6}{5} \times \binom{1}{0} = 192$
The answer I get by this approach is correct. However, there is an approach mentioned in my module which counts this whole thing in just $\binom{6}{5} \times 2^5 = 192$. But unfortunately I couldn't understand this approach. Could anybody explain ?