group 12 boys and 8 girls with restrictions 
There're 12 boys and 8 girls in a class. The teacher wants to randomly split them into 3 groups: 5 kids in group A, 11 kids in group B and 4 kids in group C. 
1) What is the probability that John and Peter will not be in the same group?
2) What is the probability that in each group there will be at least one boy?

It looks like both questions can be solved by using an event's complement.
First let $|T|$ be the total possible ways to split kids into groups. $|T| = \binom{20}{5}\binom{15}{11}\binom{4}{4}$.
1) Let the event be named $E_1$, then its complement is $E_1^c$ which is an event such that John and Peter will be in the same group. If they're both in group A then there're 18 kids left to split and 3 kids to choose in A. In such scenario there're options to split the kids which is $\binom{18}{3}\binom{15}{11}\binom{4}{4}$. If John and Peter are in group B then there're $\binom{18}{9}\binom{9}{5}\binom{4}{4}$ ways to choose and if John and Peter are together in group C then there're $\binom{18}{2}\binom{16}{11}\binom{4}{4}$ ways to choose. Hence $|E_1^c| = \binom{18}{3}\binom{15}{11}\binom{4}{4} + \binom{18}{9}\binom{9}{5}\binom{4}{4} + \binom{18}{2}\binom{16}{11}\binom{4}{4}$ and $P(E_1) = 1 - \frac{P(E_1^c)}{P(T)}$. 
2) Let the event that in each group there's at least one boy be $E_2$. Then the event such that there's not at least one boy in each group is its compliment $E_2^c$. Suppose that in A all kids are girls. There're $\binom{8}{5}\binom{15}{11}\binom{4}{4}$ ways to split the kids into groups with such restriction. Suppose that in C all kids are girls, then there're $\binom{8}{4}\binom{16}{11}\binom{4}{4}$ ways to split them. Lastly, there group B is composed of 11 kids so there will always be boys and girls in that group. Hence $|E_2^c| = \binom{8}{4}\binom{16}{11}\binom{4}{4} + \binom{8}{5}\binom{15}{11}\binom{4}{4}$ and $P(E_2) = 1 - \frac{P(E_2^c)}{P(T)}$.
The last one is really tricky so not sure I got it.
 A: The total number of ways to split them into groups is:
$$\frac{(12+8)!}{5!\times11!\times4!}=21162960$$

Question #$1$:
The number of combinations with John and Peter in the 1st group is:
$$\frac{(12+8-2)!}{(5-2)!\times11!\times4!}=1113840$$
The number of combinations with John and Peter in the 2nd group is:
$$\frac{(12+8-2)!}{5!\times(11-2)!\times4!}=6126120$$
The number of combinations with John and Peter in the 3rd group is:
$$\frac{(12+8-2)!}{5!\times11!\times(4-2)!}=668304$$
So the probability that John and Peter will not be in the same group is:
$$\frac{21162960-(1113840+6126120+668304)}{21162960}\approx62.63\%$$

Question #$2$:
The number of combinations with the 1st group consisting of girls only is:
$$\binom{8}{5}\times\frac{(12+(8-5))!}{11!\times4!}=76440$$
The number of combinations with the 3rd group consisting of girls only is:
$$\binom{8}{4}\times\frac{(12+(8-4))!}{5!\times11!}=305760$$
So the probability that there will not be a group consisting of girls only is:
$$\frac{21162960-(76440+305760)}{21162960}\approx98.19\%$$
