Question

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

Short answer

There is \(54600\)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\(n! = n \cdot (n - 1) \cdot \ldots \cdot 2 \cdot 1\)

An equal number of men and women out of \(6\) members, require that there are \(3\)men and \(3\)women and find out the product of them

Women:

\(3\)of the \(15\) women should be selected.

\(\begin{array}{l}n = 15\\r = 3\end{array}\)

Evaluate the definition of a combination:

\(\begin{array}{c}C(15,3) = \frac{{15!}}{{3!(15 - 3)!}}\\ = \frac{{15!}}{{3!12!}}\\ = 455\end{array}\)

Men:

\(3\)of the \(10\) men should be selected.

\(\begin{array}{l}n = 10\\r = 3\end{array}\)

Evaluate the definition of a combination:

\(\begin{array}{c}C(10,3) = \frac{{10!}}{{3!(10 - 3)!}}\\ = \frac{{10!}}{{3!7!}}\\ = 120\end{array}\)

Use the product rule:

\(455 \cdot 120 = 54,600\)