Question

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?

Answer 1

Hopefully you know combinatoric idea that `((n),(k))` 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 `((7),(3))` 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`

Answer 2

`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 `9xx8xx7xx6` different committees, however this will include the same combinations of people.
There are `4xx3xx2xx1` ways in which 4 people can be chosen.

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

`(9xx8xx7xx6xx5)/(4xx3xx2xx1) = 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:

`(7xx6)/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;

`(7xx6xx5)/(3xx2xx1) = 35`

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

So, `2xx35 = 70` different committees can be formed.