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?

Apr 28, 2016

$56$ 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!)

$\textcolor{w h i t e}{\text{XXXXXX}} = \frac{8 \times 7 \times 6 \times \cancel{5 \times 4 \times 3 \times 2 \times 1}}{\cancel{5 \times 4 \times 3 \times 2 \times 1} \cdot \left(3 \times 2\right)}$

$\textcolor{w h i t e}{\text{XXXXXX}} = \frac{8 \times 7 \times \cancel{6}}{\cancel{3 \times 2}}$

$\textcolor{w h i t e}{\text{XXXXXX}} = 56$

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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 $8 \times 7 \times 6$ 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 $3 \times 2 \times 1 = 6$ arrangements should be considered the same committee.

Therefore the number of committees (without internal ordering) should be $\frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$