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A person can however receive $0$ coins. Example - Number of ways of distributing $3$ coins among $2$ people such that no person receive more than $2$ coins is $2$.

My approach - For a simpler version if we do not have restriction of each person having a maximum of C coins.

We know that $$ \sum coins(\{ (c_1,...,c_M) \}) = N $$

Then Number of ways to distribute N coins among M people will be $$ {{N+M -1}\choose{M-1}} $$

I fail to understand how to include restriction of each person having having a maximum of C coins.

Shubham Chopra
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2 Answers2

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This answer is actually incorrect, but I will leave it here in case anyone makes the same mistake I did in reasoning this out. Take a look at saulspatz's response for the right answer :)

I am assuming that $N<MC$, otherwise there are zero ways.

The idea is the following: first give each person $C$ coins, and then take away $MC-N$ coins by choosing from the $M$ people with repetition allowed and order not important.

This corresponds to the number $\big(\!\!\binom{M}{MC-N}\!\!\big) = \binom{M+MC-N-1}{MC-N}$, where $\big(\!\!\binom{n}{k}\!\!\big)$ denotes multichoose.

Luke Collins
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  • If $MC-N>C$ doesn't this count cases where you take away more than one coin from a single person? – saulspatz Aug 08 '20 at 13:35
  • For example, if $M=N=3,\ C=2$ the above gives $10$, but there are $7$ ways to distribute $3$ coins among $3$ people witha limit of $2$ coins per person. Give each person $1$ coin, or give one person $2$ and another $1$. – saulspatz Aug 08 '20 at 13:46
  • @saulspatz You're right, I overlooked this. Thank you for pointing it out. – Luke Collins Aug 08 '20 at 23:04
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The accepted answer is incorrect, as I noted in a comment. There isn't a very compact formula for this, so far as I know. You can do it with generating functions. I'll use the coefficient extractor operator. $[x^n]P(x)$ mean the coefficient of $x^n$ in the polynomial or power series $P(x)$.

The number of ways to distribute $N$ coins to $M$ people, with a limit of $C$ coins per person is $$\begin{align} &[x^N](1+x+x^2+\cdots+x^C)^M\\ =&[x^N]\left(\frac{1-x^{C+1}}{1-x}\right)^M\\ =&[x^N]\sum_{k=0}^M(-1)^k\binom Mkx^{k(C+1)}(1-x)^{-M}\\ =&[x^N]\sum_{k=0}^M(-1)^k\binom Mkx^{k(C+1)}\sum_{j=0}^\infty(-1)^j\binom{-M}{j}x^j\\ =&[x^N]\sum_{k=0}^M(-1)^k\binom Mkx^{k(C+1)}\sum_{j=0}^\infty\binom{M+j-1}{j}x^j\\ =&\sum_{k=0}^M(-1)^k\binom Mk\binom{M+N-k(C+1)-1}{N-k(C+1)} \end{align},$$ with the usual convention that $\binom nm = 0$ if $m<0$ so that we can take $$0\leq k\leq\left\lfloor\frac{N}{C+1}\right\rfloor$$ and the sum becomes
$$\sum_{k=0}^{\lfloor N/(C+1)\rfloor}(-1)^k\binom Mk\binom{M+N-k(C+1)-1}{M-1}$$

The first formula comes by considering the exponent of $x$ in the $j$th term as the number of coins given to person $j$. Then the coefficient of $x^N$ in the product counts the number of ways that we actually distributed $N$ coins. The remainder comes from the binomial theorem and binomial formula. In the very last line, we pick out the value $j$ that results in an exponent of $N$.

EXAMPLE

With $M=N=3,\ C=2$, the above formula gives $$\sum_{k=0}^1(-1)^k\binom3k\binom{5-3k}{3-3k}=\binom30\binom53-\binom31\binom20=7$$ which is correct. The accepted answer gives $10$.

EDIT

This is the same formula one would get by applying the principle of inclusion and exclusion. The first term in the sum, $\binom{N+M-1}{M-1}$ is the number of ways to distribute $N$ coins to $M$ people without restriction. In the second term, we subtract $\binom M1\binom{N-(C+1)+M-1}{M-1}$, the number of ways to give one person $C+1$ coins, and then distribute the remaining $N-(C+1)$ coins. Any distributions where two people got more than $C$ coins have been subtracted twice, so we add back the third term, $\binom M2\binom{N-2(C+1)+M-1}{M-1}$, the number of ways to give two people $C+1$ coins each, and then distribute the remaining $N-2(C+1)$ coins, and so on.

saulspatz
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