Number of solutions to $x_1+x_2+...+x_p=m$

Question

Find the number of solutions of the equation $$x_1+x_2+...+x_p=m$$ such that $x_k\geq l$ for all $k$. ($m, p, l \geq 0$ are given.)

Attempt: We can firstly give every $x$ $l$, then the problem becomes finding the number of partition of $m-pl$ with the number of summands no more than $p$, so the result is ${m-pl-1 \choose p-1}+{m-pl-1 \choose p-2}+...+{m-pl-1 \choose 0}$. Is this correct? If so, can I simplify this expression?

The number of weak compositions of $m-pl$ with $p$ parts is $\binom{m-p(l-1)-1}{p-1}=\binom{m-pl+p-1}{p-1}$. — Alexander Burstein, Feb 29 '20 at 21:57

score 3 · Answer 1 · edited Feb 29 '20 at 22:03

First you have to know the basic problem of counting solutions to $x_1 + \ldots + x_p = n$ where all $x_i \ge 0$ (and integer). This has $\binom{p + n -1}{p-1}$ many solutions (using stars and bars, e.g.)

So your idea of "giving $l$ to all $k$ bins/variables is sound, so you reduce to the number of solutions to $$x_1 +\ldots +x_p = m-lp, x_i \ge 0$$

which is by the first fact $\binom{p + (m-lp) - 1}{p-1}$ so one term, not a sum.

score 1 · Accepted Answer · answered Feb 29 '20 at 23:58

If you know about generating functions, the problem is to find the coefficient of $x^m$ in the generating function $$A(x)=\left(x^l+x^{l+1}+\cdots\right)^p.$$

Note that

\begin{align} A(x)&=x^{pl}(1+x+x^2+\cdots)^p\\ &=x^{pl}\frac{1}{(1-x)^p}\\ &=x^{pl}\sum_{k=0}^\infty\binom{p+k-1}{p-1}x^k\\ &=\sum_{k=0}^\infty\binom{p+k-1}{p-1}x^{pl+k}. \end{align}

Observe that $m\geq pl$. (Otherwise the equation has no solution.). Thus, if $pl+k=m$,we have the solution $$\binom{p+m-pl-1}{p-1}.$$

Masoud · Answer 3 · 2020-02-29T22:07:43.413

0

Hint: I think $$x_1+ \cdots x_p =m$$

now drop $l$ to each $x_i$ and got

$$y_1+ \cdots y_p =m-pl$$ $y_i=x_i-l\geq 0$ now solve new equation.

edited Feb 29 '20 at 22:07

answered Feb 29 '20 at 22:00

Masoud

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Number of solutions to $x_1+x_2+...+x_p=m$

3 Answers3