Question

Suppose that a department contains\(10\)men and\(15\)women. How many ways are there to form a committee with six members if it must have more women than men?

Short answer

There is \(96460\)ways.

Step-by-step solution

Definitions of Product rule, Permutation, and Combination

Product rule: If one event can occur in\(m\)ways and a second event can occur in\(n\)ways, then the number of ways that the two events can occur in sequence is then \(m \cdot n\).

Definition permutation (order is important):

\(P(n,r) = \frac{{n!}}{{(n - r)!}}\)

Definition combination (order is not important):

\(C(n,r) = \left( {\begin{array}{*{20}{l}}n\\r\end{array}} \right) = \frac{{n!}}{{r!(n - r)!}}\)with\(n! = n \cdot (n - 1) \cdot \ldots \cdot 2 \cdot 1\)

Consider there are more than \(3\) women and less than \(3\) men and use the product rule

\(4\)women and \(2\) men:

\(4\)of the \(15\) women should be selected.

\(\begin{array}{l}C(15,4) = \frac{{15!}}{{4!(15 - 4)!}}\\C(15,4) = \frac{{15!}}{{4!11!}}\\C(15,4) = 1365\end{array}\)

\(2\)of the \(10\) men should be selected.

\(\begin{array}{l}C(10,2) = \frac{{10!}}{{2!(10 - 2)!}}\\C(10,2) = \frac{{10!}}{{2!8!}}\\C(10,2) = 45\end{array}\)

Use the product rule:

\(1365 \cdot 45 = 61425\)

Consider there are more than \(3\) women and less than \(3\) men and simplify by the product rule

\(5\)women and \(1\) man:

\(5\)of the \(15\) women should be selected.

\(C(15,5) = \frac{{15!}}{{5!(15 - 5)!}} = \frac{{15!}}{{5!10!}} = 3003\)

\(1\)of the\(10\)men should be selected.

\(C(10,1) = \frac{{10!}}{{1!(10 - 1)!}} = \frac{{10!}}{{1!9!}} = 10\)

Use the product rule:

\(3003 \cdot 10 = 30030\)

Consider there are more than \(3\) women and less than \(3\) men and use the product rule then add the values

\(6\)women and\(0\)men:

\(6\)of the\(15\)women should be selected.

\(C(15,6) = \frac{{15!}}{{6!(15 - 6)!}} = \frac{{15!}}{{6!9!}} = 5005\)

\(0\)of the\(10\)men should be selected.

\(C(10,0) = \frac{{10!}}{{0!(10 - 0)!}} = \frac{{10!}}{{0!10!}} = 1\)

Use the product rule:

\(5005 \cdot 1 = 5005\)

Use the sum rule:

\(61425 + 30030 + 5005 = 96460\)