A class consists of 3 boys and 6 girls willing to form 3 groups of 3 called Groups A, B, C. How many ways are there to assign 9 of them to Groups A, B C?
I started with $\frac{9!}{3!3!3!}$ but it seems faulty because there might be 3! ways to label each group formed as A, B, C. Not sure.
Part 2. How many ways are there to have exactly 1 group with all boys?
On
The answer $\binom{9}{3,3,3}=\frac{9!}{(3!)^3}$ that you've written down already includes this possibility; you can view it as choosing three people out of nine, and putting them in group A; of the remaining six choose three more, for group B; then the remaining three are group C. This means the count is $$ \binom{9}{3}\binom{6}{3}\binom{3}{3}=\frac{9!}{3!\cdot 6!}\cdot\frac{6!}{3!\cdot3!}\cdot\frac{3!}{3!\cdot0!}=\frac{9!}{3!\cdot3!\cdot 3!}=1680. $$ If you wanted to count the number of ways to group people up without caring about labels, you could do so by putting them in labeled groups, then dividing by the number of ways to label the groups; this would yield $$ \frac{9!}{3!\cdot3!\cdot3!\cdot3!}=280. $$
If each group needs to have a boy in them, pick the boys first: the A group has three choices, the B group has two choices, the last, C 1. Independently, pick the first and girl in each group, that's 6*5/2 , 4*3/2, 2*1/2 so $$3! \cdot 6! \over 2^3$$