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PROBLEM: How many ways can 10 identical DVDs be put in 3 identical boxes so that each box contains at least 2 DVD's.

My reasoning is this ..

Since each box must contain at least 2 DVDs, I take 3 out of the 10 DVDs out and automatically assign each of them to a box. So now all the boxes have 1 DVD, and now 7 are left.

The problem has now been reduced to how many ways can you assign 7 DVDs into 3 boxes so that each box gets at least 1 DVD. So here using stars and bars, you can use 2 bars to create 3 partitions of the 7 DVDs. Between the 7 DVDs there are 6 spots to put the 2 bars in. This yields 6 choose 2 combinations which is 15.

The correct answer is 4, Im lost, could someone please explain.

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    $\begingroup$ Why did stop at inserting just $1$ per box ? you can go straight for $2$, and then you have the number of ways to distribute $4$ DVDs into $3$ boxes, which in fact is $4$. $\endgroup$
    – G Cab
    Nov 8, 2016 at 15:19

2 Answers 2

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You had a good start except i believe you missed the part where the question said each box needs to have at least 2 DVD's.

We assign $2$ DVD's to each box and remaining 4 DVD's are distributed in $\binom43 = 4$ ways.

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You can start off the same way you mentioned, by distributing CDs into each box; instead of distributing one to each, you can distribute two to each box and simplify the problem to distributing 4 identical CDs into 3 identical boxes.

We cannot use stars and bars here, because the boxes are identical.
There are four ways to partition 4 into 3 parts:
4, 0, 0
3, 1, 0
2, 2, 0
2, 1, 1

Thus, we get 4 solutions.

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