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In how many ways can $20$ identical balls be distributed in $4$ distinct boxes, subject to the following conditions:

  1. Each box has at least $2$ balls,
  2. Each box has an even number of balls?

The distribution of $20$ identical balls in $4$ distinct boxes is equal to the sequence of $20$ $0$'s and $3$ $1$'s. So first we put $2$ balls in each box so no box is left empty( We have $1$ way of doing so), then we distribute the remaining $12$ balls in $4$ boxes(The formula is $C(k + n - 1, k),$ from the theorem: The number of $k$-combinations with unlimited repetition of a set of $n$ distinct objects is $C(k +n−1, k)$)

Which then implies $C( 12 + 4 - 1, 12)$, so the total number of sequences of $0$'s and $1$'s is $C(15,12) = 15! / (12! * 3!)

I am not sure if the above written is correct and I do not know how to find the answer for the $2^{nd}$ condition.

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You want to know how many ways there is to place 20 identical balls in 4 boxes.

Given Each box has an even number of balls, this becomes equal to the number of ways there is to place 10 pairs of balls in 4 boxes...

Given Each box has at least 2 balls, this becomes equal to placing one pair in each box first, then counting the number of ways there is to place 6 (remaining) pairs of balls in 4 boxes.

That should cover most of the distance between the problem and the solution.

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