200 identical candies are to be sent to 5 families (A, B, C, D, E).

Question

200 identical candies are to be sent to 5 families (A, B, C, D, E). If each family must get at least 3 candies and the family “A” cannot have more than 30 candies, how many different ways are there to distribute the candies?

Answer 1

Let $a_n$ be the number of ways to distribute $n$ candies to $4$ families. Then the answer is \begin{equation} \newcommand{\choose}[2]{\left(\begin{array}{c}#1\\#2\end{array}\right)} a_{170} + a_{171} + a_{197}. \end{equation} It can be shown that \begin{equation} a_n = \choose{n-9}{3}, \end{equation} hence the answer is \begin{equation} a_{12} + a_{13} + \cdots + a_{170} = \choose{161}{3} + \choose{162}{3} \cdots + \choose{188}{3}. \end{equation}

Combining my result and Iris's gives the following equation! :) \begin{equation} \choose{161}{3} + \choose{162}{3} \cdots + \choose{188}{3} = \choose{189}{4} - \choose{161}{4}. \end{equation}

Answer 2

We can give each family three candies first, so we have $200-3\times 5=185$ candies left to distribute.

Now we have total $\frac{189\times188\times187\times186}{4\times3\times2\times1}=51,494,751$ ways.

Since $A$ can’t get more than $30$ candies, we can minus the ways that $A$ get at least $31$ candies,which is:

First give $A$ $31$ candies, also $B,C,D,E$ get $3$ candies first, so we have $200-31-3\times4=157$ to distribute. So there are $\frac{161\times160\times159\times158}{4\times3\times2\times1}=26,964,280$ ways.

Hence there are $51494751-26964280=24530471$ ways.

Answer 3

By the way, in coming my previous result and Iris's result, I found the following equality!

\begin{equation} \newcommand{\choose}[2]{\left(\begin{array}{c}#1\\#2\end{array}\right)} \choose{n}{p} + \choose{n+1}{p} + \cdots + \choose{m}{p} = \choose{m+1}{p+1} - \choose{n}{p+1} \end{equation} where $m$, $n$, and $p$ are natural numbers satisfying $m\geq n\geq p$.