Distributing 10 swords among 3 champions.

Question

Right, so the protocol here seems to be stars and bars. Now let's say the blacksmith is to make sure that each champion gets two swords. We may fix two for each champion and count the combinations for the remaining $10-(3\times2)=4$ swords.

$$ \binom{4+3-1}{3-1} = 15 $$

But my first take to this "two for each" requirement yielded the use of the theorem one on the linked Wikipedia article. I thought if we distribute 5 swords to 3 champions making sure each gets at least one and then distribute again 5 swords giving each at least one then we have distributed 10 swords making sure each champion gets two.

$$ 2\times\binom{5-1}{3-1} = 12 \neq 15 $$

I am inclined to believe that 15 is the correct count but can't see why the second approach is counting under. Could someone tell what cases the second approach misses and why?

Answer 1

There are two problems:

1. If you (a) distribute 5 swords to 3 champions making sure each gets at least one and (b) do that again, the total amount of outcomes is not $\binom{5-1}{3-1} + \binom{5-1}{3-1} = 6+6 = 12$ but $\binom{5-1}{3-1} \cdot \binom{5-1}{3-1} = 6 \cdot 6 = 36$. So we should be looking for an overcount, not an undercount.
2. The reason this is bigger than $15$ is because certain outcomes are counted multiple times. If a champion gets $2$ swords in the first round and $1$ sword in the second round, that should be the same outcome as getting $1$ sword in the first round and $2$ swords in the second round, but $6 \cdot 6$ counts these separately.

We can confirm that $15$ is correct by brute force: the options are

\begin{array}{ccc} 6+2+2 & 2+6+2 & 2+2+6 \\ 5+3+2 & 3+5+2 & 3+2+5 \\ 5+2+3 & 2+5+3 & 2+3+5 \\ 4+3+3 & 3+4+3 & 3+3+4 \\ 4+4+2 & 4+2+4 & 2+4+4 \end{array}