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In how many ways can $4$ different Green balls and $4$ different Red balls be Distributed to $4$ persons equally such that each will get balls of same color.

My Try: Let Green balls be $G_1$, $G_2$, $G_3$ and $G_4$ and Red balls be $R_1$,$R_2$, $R_3$ and $R_4$.

Now each person should get $2$ Green, $2$ Green, $2$ Red, $2$ Red Respectively.

Number of ways of dividing $4$ Green balls into two groups of $2$ each is $$\frac{\binom{4}{2}}{2}=3$$ Similarly Number of ways of dividing $4$ Red balls into two groups of $2$ each is $3$.

If we concatenate each green and red grouping for example:

$G_1G_2$ $G_3G_4$ $R_1R_2$ $R_3R_4$..so each of these $4$ groupings can be arranged in $4!$ ways for four persons. But total number of concatenations is $9$. So Number of ways is $$9.4!=216$$.

Can i know any other approach

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Line up the people in order of student number. The greens people can be chosen in $\binom{4}{2}$ ways. For each such way the leftmost greens person can choose her greens in $\binom{4}{2}$ ways. Then the leftmost reds person can choose her reds in $\binom{4}{2}$ ways, for a total of $\binom{4}{2}^3$.

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