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I know that we can distribute $n$ chocolates among $k$ children in ${n+k-1 \choose k-1}$ ways which is the stars and bars problem. But what if the chocolates should be given in increasing order? I mean the 1st boy gets less chocolates than the second boy and so on.

If $a_1+a_2+a_3+\cdots+a_k=n$, then in how many ways can we distribute $n$ such that $a_1<a_2<a_3<\cdots<a_k$?

If such a numerical problem is given, we can reduce $n$ to a smaller number and build various cases to solve it. But is there some generalized method to solve this kind of problems?

tomriddle99
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    Let $x_1=a_1, x_2=a_2-a_1, x_3 = a_3-a_2,\dots, x_k=a_k-a_{k-1}$. We equivalently have that $kx_1+(k-1)x_2+(k-2)x_3+\dots+2x_{k-1}+x_k=n$, but now the only restriction is $x_1\ge 0$ and $x_i>0$ for $i=2,\dots,k$. With a little more finagling, that reduces to this previously asked question: https://math.stackexchange.com/questions/2053319/number-of-solutions-of-x-12x-2-cdotskx-k-n. – Mike Earnest May 10 '19 at 03:55
  • $<$ (strictly increasing) or $\le$ (in order)? – Peter Taylor May 10 '19 at 12:23
  • @PeterTaylor, strictly increasing. – tomriddle99 May 10 '19 at 14:33

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The search term you need is partitions of $n$ into $k$ distinct parts".

The easy way to calculate them is a 3-variable recursion with $n$, $k$, and an upper bound on the largest part.

As a starting point for a literature search you could take https://core.ac.uk/download/pdf/82124189.pdf

Peter Taylor
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