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How many ways are there to pack $9$ identical DVDs into $3$ indistinguishable boxes so that each box contains at least two DVDs?


Is this having one to one correspondence with the partitions of a number?

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    $\begingroup$ Put two DVD's in each box. You still have $3*3=9$ ways of placing the remaining 3 DVD's. $\endgroup$ – Evargalo Jul 28 '17 at 13:17
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Yes, it's the number of partitions of $3$ with at most $3$ parts, but since $3\leq3$ the extra condition doesn't add anything in this case.

(The first $3$ is the number of DVDs you have left after putting the minimum number into each box, and the second is the number of boxes.)

So there are $3$ ways, which correspond to the partitions $3$, $2+1$, $1+1+1$, and have boxes with $5,2,2$ or $4,3,2$ or $3,3,3$ DVDs respectively.

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  • $\begingroup$ Thanks, This is what I wanted !! $\endgroup$ – Jon Garrick Jul 28 '17 at 13:29

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