# Distributing indistinguishable coins into distinguishable children

Abe wants to distribute 50 (indistinguishable) coins among his 12 (distinguishable) grandchildren, in such a way that all three of his children, A, B, C gets at least 3 coins each. In how many ways can Abe do this?

My solution begins by distributing three coins to A, B, C, and then use starts and bars to distribute the rest:

Without loss of generality, let's begin distributing to A, B, and C in that order.

There are $$\binom{50}{3}$$ ways to distribute to A.

There are $$\binom{47}{3}$$ ways to distribute to B.

There are $$\binom{44}{3}$$ ways to distribute to C.

Then there are 41 coins to be distributed among 12 grandchildren, so using stars and bars with the formula: $$\binom{n+r-1}{r-1}$$, where $$n$$ denotes the number of indistinguishable items to distribute, and $$r$$ denotes the bins.

We have $$\binom{41+12-1}{12-1} = \binom{52}{11}$$ ways to distribute the rest of the coins.

In total, we have $$\binom{50}{3}\binom{47}{3}\binom{44}{3}\binom{52}{11}$$ ways to distribute.

However, I got the question wrong, and the notes says "Not accounting for "at least 3 coins each" condition correctly".

I am wondering what was I missing here?

There are $$50$$ identical coins. So you do not have $${50 \choose 3}$$ ways to choose coins for grandchild $$A$$, for example.

What you are finding is positive integer solutions (and including zero) to

$$a_1 + a_2 + ...+ a_{12} = 50$$ with the constraint $$a_1, a_2, a_3 \geq 3$$.

This is equivalent to finding solutions to

$$a_1 + a_2 + ...+ a_{12} = 41$$ without constraint.

$$\big[$$Think of it as giving $$3$$ identical coins each to $$a_1, a_2, a_3$$ first and there is just one way to do that; then distributing rest $$41$$ to all $$12$$ grandchildren without constraint$$\big]$$.

• So the answer is just stars and bars for 41 indistinguishable items to 12 urns? Mar 8, 2021 at 19:51
• Yes that is correct. Mar 8, 2021 at 19:53