We have $3$ groups $A,B,C$.
We have to divide all the boys and the girls into these three groups such that in group $C$, there's exactly $6$ or $7$ members, there's atleast one girl, and the number of the boys must be greater than the number of girls.
in how many ways we can do that?
My attempt: I'm trying to split this problem into cases:
Case 1: in group $C$ theres $5$ boys and $1$ girl, or $6$ boys and $1$ girl. for the rest of the boys and girls I would have to choose a group for each one from $A,B$ so we get $2^{14}+2^{13}$.
Case 2: in group $C$ theres $4$ boys and $2$ girls or $5$ boys and $2$ girls. so $2^{14}+2^{13}$.
Case 3: in group $C$ theres $4$ boys and $3$ girls: $2^{13}$.
Then I summed up all the results and got $57344$ ways and the answer was $50610$.
What am I missing?
Appreciate any help, thanks in advance.
Best Answer
It seems that the question is only about the possible combinations for group $C$.
This is where the answer 50610 comes from:
$$\binom{10}{5}\cdot\binom{10}{1} + \binom{10}{6}\cdot\binom{10}{1} + \binom{10}{4}\cdot\binom{10}{2} + \binom{10}{5}\cdot\binom{10}{2} + \binom{10}{4}\cdot\binom{10}{3} = 50610$$
For example: to choose 5 boys (out of 10) and 1 girl (out of 10) to put into group $C$ we have $\binom{10}{5}\cdot\binom{10}{1}$ possibilities.