Question

How many ways can n identical balls be distributed into k bins such that each bin contains at least two balls? Assume that n ≥ 2k.

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Answer 1

The number of ways of arranging $\fn_jvn n$ identical balls in $\fn_jvn m$ distinct urns is the number of terms in the expansion of $\fn_jvn (x_1+x_2+...+x_m)^n$ . This is $\fn_jvn \binom{n+m-1}{n}$ .

Here $\fn_jvn k=m$ and we need at least 2 ball in in a bin. Distribute 2 balls in each of $\fn_jvn k$ different bins. So consider distributing $\fn_jvn n-2k$ identical balls in $\fn_jvn k$ different bins.

The number of possible ways is

the number of ways of arranging $\fn_jvn n-2k$ identical balls in $\fn_jvn k$ distinct bins is the number of terms in the expansion of $\fn_jvn (x_1+x_2+...+x_k)^{n-2k}$ . This is $\fn_jvn \binom{n-2k+k-1}{n-2k}={\color{Blue} \binom{n-k-1}{n-2k},n\geqslant 2k}$ .