In how many ways can 12 balls be distributed | vedclass

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(B) Let the two friends be $F_1$ and $F_2$. Case $1$: $F_1$ gets $8$ balls and $F_2$ gets $4$ balls. The number of ways is $\binom{12}{8} 	imes \bino

Vedclass · Accepted Answer

$\frac{2!12!}{8!4!}$

Vedclass · Answer

(B) Let the two friends be $F_1$ and $F_2$. Case $1$: $F_1$ gets $8$ balls and $F_2$ gets $4$ balls. The number of ways is $\binom{12}{8} 	imes \binom{4}{4} = \frac{12!}{8!4!}$. Case $2$: $F_1$ gets $4$ balls and $F_2$ gets $8$ balls. The number of ways is $\binom{12}{4} 	imes \binom{8}{8} = \frac{12!}{4!8!}$. Since these are two distinct cases,the total number of ways is $\frac{12!}{8!4!} + \frac{12!}{4!8!} = 2 	imes \frac{12!}{8!4!} = \frac{2! 	imes 12!}{8!4!}$.

In how many ways can $12$ balls be distributed between two friends such that one gets $8$ balls and the other gets $4$ balls?

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