Chapter 4.4, Problem 43E

Selecting a Committee There are 7 women and 5 men in a department. How many ways can a committee of 4 people be selected? How many ways can this committee be selected if there must be 2 men and 2 women on the committee? How many ways can this committee be selected if there must be at least 2 women on the committee?

To determine

The number of ways to select 4 people for the committee, 2 men and 2 women for the committee and at least 2 women for the committee.

Answer to Problem 43E

There are 495 ways to select 4 people out of 12 people.

There are 210 ways to select 2 men and 2 women

There are 420 ways to select at least 2 women.

Explanation of Solution

Given info:

A committee must be formed with 4 members. There are 7 women and 5 men available for the selection.

Calculation:

Combinations:

A combination is an arrangement of n objects in r ways where the order of arrangement is not considered.

${}_n{C}_r=\frac{n!}{\left(n-r\right)!r!}$

Where, n is the total number of objects and r is number of ways in which n objects can be selected.

The number of ways to select 4 people out of 12 people:

Substitute n as 12 and r as 4.

$\begin{array}{c}{}_{12}{C}_4=\frac{12!}{\left(12-4\right)!4!}\\ =\frac{12\times 11\times 10\times 9\times \cancel{8!}}{\cancel{8!}4!}\\ =\frac{11,880}{24}\\ =495\end{array}$

Thus, there are 495 ways to select 4 people out of 12 people.

The number of ways to select 2 women and 2 men out of 7 women and 5 men:

Substitute n as 7 and r as 2.

$\begin{array}{c}{}_7{C}_2=\frac{7!}{\left(7-2\right)!2!}\\ =\frac{7\times 6\times \cancel{5!}}{\cancel{5!}2!}\\ =\frac{42}{2}\\ =21\end{array}$

Thus, there are 21 ways to select 2 women out of 7 women.

Substitute n as 5 and r as 2.

$\begin{array}{c}{}_5{C}_2=\frac{5!}{\left(5-2\right)!2!}\\ =\frac{5\times 4\times \cancel{3!}}{\cancel{3!}2!}\\ =\frac{20}{2}\\ =10\end{array}$

Thus, there are 10 ways to select 2 men out of 5 men.

Fundamental counting rule:

The number of ways in which a sequence of n events occur if the first event can occur in $k_1$ ways, the second event can occur in $k_2$ ways and so on.

$k_1\times k_2\times \mathrm{...}\times k_n$

The number ways to select 2 men and 2 women is given below:

$\begin{array}{c}{}_7{C}_2{\times }_5{C}_2=21\times 10\\ =210\end{array}$

Thus, there are 210 ways to select 2 men and 2 women.

The number of ways to select at least 2 women for the committee:

There can be 2 women, 3 women and 4 women in the committee.

The number of ways to select 2 women and 2 men is 210 ways.

The number of ways to select 3 women and 1 man

Substitute n as 7 and r as 3.

$\begin{array}{c}{}_7{C}_3=\frac{7!}{\left(7-3\right)!3!}\\ =\frac{7\times 6\times 5\times \cancel{4!}}{\cancel{4!}3!}\\ =\frac{210}{6}\\ =35\end{array}$

Thus, there are 35 ways to select 3 women out of 7 women.

Substitute n as 5 and r as 1.

$\begin{array}{c}{}_5{C}_1=\frac{5!}{\left(5-1\right)!1!}\\ =\frac{5\times \cancel{4!}}{\cancel{4!}1!}\\ =5\end{array}$

The number ways to select 2 men and 2 women is given below:

$\begin{array}{c}{}_7{C}_3{\times }_5{C}_1=35\times 5\\ =175\end{array}$

Thus, there are 175 ways to select 3 women and 1 man.

The number of ways to select 4 women and 0 men

Substitute n as 7 and r as 4.

$\begin{array}{c}{}_7{C}_4=\frac{7!}{\left(7-4\right)!4!}\\ =\frac{7\times 6\times 5\times \cancel{4!}}{\cancel{4!}3!}\\ =\frac{210}{6}\\ =35\end{array}$

Thus, there are 35 ways to select 4 women out of 7 women.

The number of ways to select at least 2 women for the committee is given below:

$210+175+35=420$

Thus, there are 420 ways to select at least 2 women for the committee.