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How many ways to distribute 11 identical balls into 3 identical boxes with each box having 2 balls at least

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closed as off-topic by Shailesh, Watson, hardmath, Henrik, Namaste Jul 25 '16 at 18:23

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    $\begingroup$ Any thoughts? Hint: Since each box has at least $2$, you really only have $5$ balls to distribute. How many ways can $5$ be written as the sum of three non-negative integers? $\endgroup$ – lulu Jul 25 '16 at 12:28
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    $\begingroup$ I assume you want to know the probability of each box having at least 2 balls? Or do you just want to know how many ways that can happen? Have you tried anything? As written, your "question" is likely to get closed unless you add more details. $\endgroup$ – jdods Jul 25 '16 at 12:30
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As each box must have at least two balls, let them first be filled with two balls each. Then, we have five balls left.

The ways in which five can be expresses as the sum of three non-negative integers (where order does not matter) are: $$5+0+0$$ $$4+1+0$$ $$3+2+0$$ $$3+1+1$$ $$2+2+1$$

Each represents a way in which the remaining balls can be put into the three boxes. Since the boxes are identical, each way will be counted once and the order doesn't matter.

So, number of ways$=5$.

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