# Distributing $4$ different red balls and four different green balls to $4$ people

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?

• Could you please explain a little bit about distribution theorem? And why the denominator is $132$? – Ning Wang Apr 19 '18 at 2:21
• Certainly $m \neq \frac{8!\cdot 4!}{2!\cdot2!\cdot2!\cdot2!\cdot2^4},$ so the application of the "distribution theorem," whatever that is, appears to have been done incorrectly. – David K Apr 19 '18 at 2:39
• How to apply distribution theorem in this question. – Pravin Kumar Apr 20 '18 at 13:18

$$\textrm{(who get green)}\cdot\textrm{(which two for each whom)}$$ $${4\choose2}\cdot({4\choose2}{2\choose2}\cdot{4\choose2}{2\choose2}),$$ which is $6\cdot6\cdot6=216.$
$$\textrm{(deliver red ones)}\cdot\textrm{(deliver green ones)}$$ $$4!\cdot4!=24\cdot24=576.$$
so the answer could be $\frac{216+576}{132}=6.$