Question

In how many ways can a committee of 4 be selected from 6 men and 8 women if the committee must contain at most 2 women?

Answer 1

595

Explanation:

We have a committee that will have 4 people and at most 2 can be women. There are 6 men and 8 women that can be on the committee. How many ways can we select the 4 people?

There are two ways we can do this - we can figure out the number of ways the committee can be formed with 0, 1, and 2 women and add them up. The other way to do it is to figure out how many different ways the committee can be formed in all cases, then subtract out the ways that involve 3 and 4 women. Let's do it both ways to show that the answer will be the same.

All of these calculations will be Combinations (we don't care about the order of the picks, just the members of the committee). The general formula for a combination is to look at P, the population (or available number of people who can sit on the committee) and k, the number selected), with the general formula being:

`C_(P,k)=(P!)/((k!)(P-k)!)`

Adding up:

0 Women

There are 6 men and 4 slots, so we have a combination of:

`C_(6,4)=(6!)/((4!)(2!))=(6xx5xx4!)/(2xx4!)=15`

1 Women

There are 8 women that can go into the 1 slot:

`C_(8,1)=(8!)/((1!)(7!))=(8xx7!)/(7!)=8`

and 6 men who can go into the remaining 3 slots:

`C_(6,3)=(6!)/((3!)(3!))=(6xx5xx4xx3!)/(3xx2xx3!)=20`

which gives us:

`8xx20=160`

2 Women

There are 8 women that can go into 2 slots:

`C_(8,2)=(8!)/((2!)(6!))=(8xx7xx6!)/(2xx6!)=28`

and 6 men who can go into the remaining 2 slots:

`C_(6,2)=(6!)/((2!)(4!))=15` (see the 0 women category for the calculation)

which gives us:

`28xx15=420`

So all told:

`15+160+420=595`

Subtracting down

The total number of ways we can fill the slots on the committee is:

`C_(14,4)=(14!)/((4!)(10!))=(cancel(14)^7xx13xxcancel(12)xx11xx10!)/(cancel(4xx3)xxcancel(2)xx10!)=7xx13xx11=1001`

4 Women

There are 8 women that can go into 4 slots:

`C_(8,4)=(8!)/((4!)(4!))=(cancel(8)^2xx7xxcancel6xx5xx4!)/(cancel4xxcancel(3xx2)xx4!)=70`

3 Women

There are 8 women that can go into 3 slots:

`C_(8,3)=(8!)/((3!)(5!))=(8xx7xx6xx5!)/(3xx2xx5!)=56`

and 6 men who can go into the remaining 1 slot:

`C_(6,1)=(6!)/((1!)(5!))=(6xx5!)/(5!)=6`

which gives us

`56xx6=336`

So all told:

`1001-70-336=595`