Combinatorics and Probability 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 them to the groups such that exactly 1 group has no boys?
 A: The label of the no boy group can be chosen in $3$ ways. For each such way, the girls who will be in that group can be chosen in $\binom{6}{3}$ ways.
Now consider the group lowest in the alphabet which will have at least one boy. That group can have (1) $1$ boy or (ii) $2$ boys.
To count (i), the boy can be chosen in $\binom{3}{1}$ ways, and for each choice of boy the $2$ girls can be chosen in $\binom{3}{2}$ ways, for a total of $\binom{3}{1}\binom{3}{2}$.
The same argument shows that the count of (ii) is $\binom{3}{2}\binom{3}{1}$.
That gives a total of $18$. Multiply by $3\binom{6}{3}$.
Because there are many ways to count, we do the arithmetic. The total number of ways is $(3)(20)(18)$, which is $1080$.  
Remark: The number $1080$ agrees numerically with the number you gave in a comment. Since reasons of space precluded giving a justification, it is not clear whether the reasoning was correct. 
A: Suppose A is the group without boys. A choice of three girls must be made from a total of six. So the number of possible assignments for group A is $\binom{6}{3}$. Then the number of assignments for group B is $\binom{6}{3}$ (since three girls are in group A, there are six people remaining to choose from). And there is only one way to assign the remaining three to group C. So there are $\binom{6}{3}^2$ ways to assign them to groups, assuming that A has no boys.
Make the same assumption (that the group has no boys) for B and then C. This means that there are 3$\binom{6}{3}^2$ ways to assign them to the groups with the given condition. 
Note: this may not be correct, but hopefully it is helpful.
A: *

*There are $6$ girls, out of which we need to necessarily form a group with $3$ girls. That can be done in $6\text{C}_3$ ways. 

*Now that we've finished the no-boys group, the remaining $2$ groups have $3$ members each out of which $3$ are guys and $3$ are girls. We need to have at least $1$ girl in both the remaining groups so that no group other than the one which we already chose has all $3$ girls. This can be done in the following manner: You need to select select a girl for the remaining $2$ groups for sure. That can be done in $3\times2$ ways. 

*Now we can select the rest of the 4 people can be put in the 2 groups in $4\text{C}_2$ ways. Hence the answer is:$$6\text{C}_3 \times3\times2\times4\text{C}_2=720$$

