# How many 4 person committees are possible from a group of 9 people if ?

## A) there are no restrictions? B) Both John and Barbara must be on the committee? C) Either John or Barbara (but not both) must be on the committee?

Apr 17, 2017

Hopefully you know combinatoric idea that $\left(\begin{matrix}n \\ k\end{matrix}\right)$ means "n-choose-k".

(a)

The number of different ways of choosing 4 items from a group of 9 items is simply: ((9),(4)) = (9!)/(5!4!) = 126

(b)

You have already chosen 2 of the committee so you are now looking at choosing the remaining 2 from a group of 7:

((7),(2)) = (7!)/(5! 2!) =21

(c)

For all committees that has John onboard, you need to choose 3 more, but you are choosing from a group of 7 as Barbara cannot be on the same committee. So in that scenario you have $\left(\begin{matrix}7 \\ 3\end{matrix}\right)$ ways of doing it.

The same applies if Barbara is to be on the committee and John excluded.

So, overall, the total number of ways to do this is:

2 xx ((7),(3)) = (2 xx 7!)/(4!3!) = 70

Apr 17, 2017

A) 126 different committees
B) 21 different committees
C) 70 different committees

#### Explanation:

Recall: The number of different ways of arranging $n$ numbers is n!

If 4 people are to be selected from 9 people:
There are 9 different choices for the first person.
There are 8 different choices for the second person.
There are 7 different choices for the third person.
There are 6 different choices for the fourth person.

This gives $9 \times 8 \times 7 \times 6$ different committees, however this will include the same combinations of people.
There are $4 \times 3 \times 2 \times 1$ ways in which 4 people can be chosen.

A) Therefore, for 4 people chosen from 9 there are

$\frac{9 \times 8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 126$ different committees.

B) If two people must stand, there are 2 people to be chosen from the remaining 7. Applying the same thinking as above this gives:

$\frac{7 \times 6}{2} = 21$ different committees.

C) if either one or the other must stand there are then 3 people who must be chosen from the remaining$7$ people;

$\frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

But this can happen in 2 different ways, depending on whether John or Barbara are on the committee.

So, $2 \times 35 = 70$ different committees can be formed.