Question

From 8 names on a ballot, a committee of 3 will be elected to attend a political national convention. How many different committees are possible?

Answer 1

`56` different committee are possible

Explanation:

This is a "how many ways can you choose `n` items from a pool of `m` items?" type of question.

The general form of the answer is "`m` Choose `n`";
written as:
`color(white)("XXX")mCn=(m!)/((m-n)!n!)`

In this case
`color(white)("XXX")8C3=(8!)/(5!*3!)`

`color(white)("XXXXXX")=(8xx7xx6xxcancel(5xx4xx3xx2xx1))/(cancel(5xx4xx3xx2xx1)*(3xx2))`

`color(white)("XXXXXX")=(8xx7xxcancel(6))/(cancel(3xx2))`

`color(white)("XXXXXX")=56`

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Another way to see this:

There are `8` people that could be our first choice;
after this there would be `7` people for our second choice;
and then `6` people for our third choice.
This would give us `8xx7xx6` ways of making first, second, and third choices.

But the same three people could form our 3 person committee in multiple ways. e.g. {Alice, Bob, Carol} and {Bob, Carol, Alice} should be considered as the same committee.

How many ways could 3 people be arranged among 3 sequential positions that should be treated as the same committee?
There are `3` possibilities for the first position; `2` for the second; and `1` for the third. So `3xx2xx1=6` arrangements should be considered the same committee.

Therefore the number of committees (without internal ordering) should be `(8xx7xx6)/(3xx2xx1) =56`