The number of ways to choose a smaller group from a larger group can be found using either a permutation or a combination. When the order matters, use a permutation and when the order doesn't matter, use a combination.

The order of the 6 students that are chosen doesn't matter so you need to use a combination.

The formula for a combination is \(\displaystyle{n}{C}{r}={n}\frac{!}{{{r}!{\left({n}-{r}\right)}!}}\) where n is the total number of objects in the group and rSince there are a total of n=9 students and r=6r=6 students are being chosen, you need to find \(\displaystyle{9}{C}{6}={9}\frac{!}{{{6}!{\left({9}−{3}\right)}!}}={9}\frac{!}{{6}}!{3}!\) is the number of objects being chosen.

The definition of a factorial is \(\displaystyle{n}\ne{n}⋅{\left({n}−{1}\right)}⋅{\left({n}−{2}\right)}⋯{\left({2}\right)}{\left({1}\right)}\). The factorials can then be expanded to:

\(\displaystyle{9}{C}{6}={9}\frac{!}{{6}}!{3}\ne\frac{{{9}\cdot{8}\cdot{7}\cdot{6}\cdot{5}\cdot{4}\cdot{3}\cdot{2}\cdot{1}}}{{{6}\cdot{5}\cdot{4}\cdot{3}\cdot{2}\cdot{1}}}{\left({3}\cdot{2}\cdot{1}\right)}\)

Cancelling the common factors in the numerator and denominator then given: 9C6=... Simplify=\(\displaystyle{3}\cdot{4}\cdot{7}\)

Multiplying then given 9C6=84 ways.

The order of the 6 students that are chosen doesn't matter so you need to use a combination.

The formula for a combination is \(\displaystyle{n}{C}{r}={n}\frac{!}{{{r}!{\left({n}-{r}\right)}!}}\) where n is the total number of objects in the group and rSince there are a total of n=9 students and r=6r=6 students are being chosen, you need to find \(\displaystyle{9}{C}{6}={9}\frac{!}{{{6}!{\left({9}−{3}\right)}!}}={9}\frac{!}{{6}}!{3}!\) is the number of objects being chosen.

The definition of a factorial is \(\displaystyle{n}\ne{n}⋅{\left({n}−{1}\right)}⋅{\left({n}−{2}\right)}⋯{\left({2}\right)}{\left({1}\right)}\). The factorials can then be expanded to:

\(\displaystyle{9}{C}{6}={9}\frac{!}{{6}}!{3}\ne\frac{{{9}\cdot{8}\cdot{7}\cdot{6}\cdot{5}\cdot{4}\cdot{3}\cdot{2}\cdot{1}}}{{{6}\cdot{5}\cdot{4}\cdot{3}\cdot{2}\cdot{1}}}{\left({3}\cdot{2}\cdot{1}\right)}\)

Cancelling the common factors in the numerator and denominator then given: 9C6=... Simplify=\(\displaystyle{3}\cdot{4}\cdot{7}\)

Multiplying then given 9C6=84 ways.