Different ways of counting the same combination What’s wrong with this reasoning:
How many different ways to pick a team of 3 from 4 people? ${4 \choose 3}$
Alternative way of counting is to choose a team of 2 from the 4 first (${4 \choose 2}$ ways), and then choose a third person (2 people remaining so 2 ways). Then, since order doesn’t matter (i.e., we could have chosen the single person first, and then the team of two) so divide by $2!$. But ${4 \choose 2} \times \frac{2}{2!} \neq {4 \choose 3}$.
 A: It's ok to think about choosing 2 people first, then a third person. But, care must be taken to avoid double counting. For example, suppose we have people $\{A, B, C, D\}$. We pick $A$ and $B$ as the initial team of two, leaving $C$ and $D$ as the remaining choices. This gives possibilities: $\{A, B, C\}$ and $\{A, B, D\}$ for our 3-person team. But, note that each of these teams could also have resulted from a different pick of the initial two person team. For example, $\{A,B,C\}$ would also be produced by picking $A$ and $C$ followed by $B$, or $B$ and $C$ followed by $A$.
In general, suppose we have $n$ people and want to pick a $k$ person team. We do this by first picking $c < k$ people. There are $\binom{n}{c}$ to do this. This leaves $n-c$ people, from which we must pick $k-c$ people to complete the team. There are $\binom{n-c}{k-c}$ ways to do this. Multiplying these numbers, we have $\binom{n}{c} \binom{n-c}{k-c}$ possibilities. But, as above, many of these possibilities will be identical, and we must avoid counting them multiple times. The number of times a given set of $k$ people appear in our list of possible teams is $\binom{k}{c}$, because this is the number of ways that those particular people could have been picked according to our procedure (i.e. by first picking $c$ of them). So, we divide by this number to obtain the number of unique possible teams:
$$\frac{\binom{n}{c} \binom{n-c}{k-c}}{\binom{k}{c}}$$
Writing out the expression shows that it's equivalent to $\binom{n}{k}$.
