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


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 $\binom{4}{2}^3$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.