Answer: (a) \(252\), (b) \(186\), (c) \(184\).

(a) There are \(10\) people in total. The number of ways to choose \(5\) people is

\({}^{10}C_5=252.\)

(b) At least \(2\) men means choosing \(2\), \(3\) or \(4\) men.

Choose \(2\) men and \(3\) others:

\({}^4C_2{}^6C_3=6\cdot20=120.\)

Choose \(3\) men and \(2\) others:

\({}^4C_3{}^6C_2=4\cdot15=60.\)

Choose \(4\) men and \(1\) other:

\({}^4C_4{}^6C_1=1\cdot6=6.\)

So the total is

\(120+60+6=186.\)

(c) Start with all selections and subtract selections missing one of the required groups.

The number with no men is

\({}^6C_5=6.\)

The number with no women is also

\({}^6C_5=6.\)

The number with no children is

\({}^8C_5=56.\)

No overlap is possible between these excluded cases, because a selection of \(5\) cannot be made from only one of the groups. Therefore the required number is

\(252-6-6-56=184.\)