Find the number of ways to form a team of four players for a game out of a group of 12 individuals

We can also organize the count in a different way. First line up the people, say in alphabetical order, or in student number order, or by height.

The first person in the lineup chooses the $3$ people (from the remaining $11$) who will be on her team. Then the first person in the lineup who was not chosen chooses the $3$ people (from the remaining $7$) who will be on her team. The double-rejects make up the third team.

The first person to choose has $\binom{11}{3}$ choices. For every choice she makes, the second person to choose has $\binom{7}{3}$ choices, for a total of $$\binom{11}{3}\binom{7}{3}.$$

Remark: The lineup is a device to avoid multiple-counting the divisions into teams. The alternate (and structurally nicer) strategy is to do deliberate multiple counting, and take care of that at the end by a suitable division.

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