In how many ways can we distribute 6 identical pears between 3 children so that each child receives at least one pear?
I am not too sure.
I thought, 6 ways to distribute to first, 5 ways to second, 4 to third.
But that gives $120$. ??
1 Answer
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6
This is a classic Stars and Bars problem, where we have $6$ indistinguishable items and $3$ distinguishable bins. Thus we have that the number of combinations is:
$$N=\binom{6-1}{3-1}=\binom{5}{2}=10$$
So there are only $10$ ways to distribute the pears such that each child has at least one pear.
answered Feb 3, 2015 at 13:22
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$\begingroup$ I agree on $\binom52$, but why do you say you have 6 items? Three of the pears are used to make sure each child has at least one, so there are only 3 to distribute. And in $\binom52$ the $5$ is 3 pears plus 2 separators. $\endgroup$ Commented Feb 3, 2015 at 13:24
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$\begingroup$ I thought in stars and bars, we start with the actual sum, which is 6. I believe @shaktal is correct, I may be wrong though. $\endgroup$– RickCommented Feb 3, 2015 at 13:25
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$\begingroup$ @HenningMakholm There are $6$ pears, and so we have $5$ places in which we can place the dividers, and we wish to place $2$ dividers in order to create three separate multisets, thus giving us $\binom{5}{2}$ $\endgroup$ Commented Feb 3, 2015 at 13:26
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$\begingroup$ @Shaktal, can I ask a question please?So are you using the form: $\binom{n + k - 1}{k}$ Or this form: $\binom{k-1}{n-1}$ where $n= 3$ and $k = 6$? The first one does not work here does it? $\endgroup$– RickCommented Feb 3, 2015 at 13:51
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$\begingroup$ @Rick I am using the second form, with $n=3$, $k=6$; as we are using Theorem 1 on the Wikipedia page, as we want positive numbers of items in each bin, rather than non-negative (which include zero). $\endgroup$ Commented Feb 3, 2015 at 13:53