Let $m$ denote the number of ways in which $4$ different balls of green colour and $4$ different balls of red colour can be distributed equally among $4$ persons if each person has balls of the same colour and $n$ be the corresponding figure when all the four persons have balls of different colours. Find $\frac{m+n}{132}$.
I've tried it using distribution theorem $\dfrac{m}{132} = \dfrac{8!}{2!\cdot2!\cdot2!\cdot2!\cdot2^4}\cdot\dfrac{4!}{132}$ but while calculating $\dfrac{n}{132}$ I am getting answer in decimal. Please help me in this?
I don't know the theorem, but I'll give it a try:
m:
$$\textrm{(who get green)}\cdot\textrm{(which two for each whom)}$$ $${4\choose2}\cdot({4\choose2}{2\choose2}\cdot{4\choose2}{2\choose2}),$$ which is $6\cdot6\cdot6=216.$
n:
$$\textrm{(deliver red ones)}\cdot\textrm{(deliver green ones)}$$ $$4!\cdot4!=24\cdot24=576.$$
so the answer could be $\frac{216+576}{132}=6.$