Basic Combinatorics I have a basic combinatorics question I am unsure how to complete, the question is as follows:
A company has 9 people in Office A, 6 in Office B and 3 in Office C. A new team of 6 people is to be formed.
How many ways can the new team be formed if:
a) The team includes two members from each office
b) Office A is to have at least two representatives
If anyone can help me with how to answer this I would be most grateful
 A: For part a), you have ${9 \choose 2} {6 \choose 2} {3 \choose 2}$. For part (b), count the number of ways to form a $6$ member team with no one from office A, which is ${9 \choose 6}$.  Count the number of ways to form a $6$ member team with exactly one person from team A, which is $9\times {9 \choose 5}$, so the the number of ways to form a $6$ member team with at least two people from A is
$${18 \choose 6} - {9 \choose 6} - 9\times{9 \choose 6}$$
A: a) There are $\binom{9}{2}$, $\binom{6}{2}$, and $\binom{3}{2}$ ways to choose $2$ persons from office $A, B, C$ respectively. So there are $\binom{9}{2}\cdot \binom{6}{2}\cdot \binom{3}{2}$ ways of choosing $6$-person teams with $2$ members from each office.
b) If $A$ has $2$ members, then the other $4$ members are chosen from $B$ and $C$, and there are $9$ from $B$ and $C$ combined. So we can have $\binom{9}{4}$ choices for the $4$ persons to form $6$-person teams. So we have $\binom{9}{2}\cdot \binom{9}{4}$ choices for for this case. If $A$ has $3$ members, then similarly we have: $\binom{9}{3}\cdot \binom{9}{3}$ choices to make $6$-person teams. And continue this way until $A$ has $6$ members, then we have the total choices is: $\binom{9}{2}\cdot \binom{9}{4} + \binom{9}{3}\cdot \binom{9}{3} + \binom{9}{4}\cdot \binom{9}{2} + \binom{9}{5}\cdot \binom{9}{1} + \binom{9}{6}$.
