At a farm auction, various pieces of farm equipment and livestock are bid on and sold.
Today, there are $7$ distinct items up for auction, and there are $3$ farmers bidding on them. One of the farmers declares that he will bid on at least $2$ of the items (not specifying which ones), and he won't be outbid.
If this farmer is telling the truth, then how many ways can the items be distributed among farmers?
Assume that all items are sold.
My solution :
$$\binom{7}{5} \cdot 3^{5}$$
Choose $5$ out of $7$ items (the other two stay with the farmer) and multiply by the number of ways of distributing $5$ distinct items to $3$ distinct people.
It's wrong and I am over counting, but I don't understand how?
The correct solution goes as follows. One bidder buys at least two goods. You can look at this as if this bidder chooses an amount of $k\in\{2,3,4,5,6,7\}$ goods, and the rest is distributed among the other two bidders. So given that bidder $1$ buys $k$ goods, for which there are $\binom{7}{k}$ possibilities, there are $2^{7-k}$ possibilities for the distribution of the $7-k$ goods among the last two bidders. This yields $$\begin{align}\sum_{k=2}^7\binom{7}{k}2^{7-k}&=\binom{7}{2}2^5+\binom{7}{3}2^4+\binom{7}{4}2^3+\binom{7}{5}2^2+\binom{7}{6}2^1+\binom{7}{7}2^0\\&=1611\neq5103=\binom{7}{2}3^5\end{align}$$ possible distributions of the $7$ goods.
So, why did you double count? This is because when you calculate $\binom{7}{5}3^5=\binom{7}{2}3^5$, you count certain possible distributions multiple times. These are exactly the distribution where the first bidder buys more than two goods, since you count distributions where bidder $1$ buys goods $1,2$ and $3$ (for example) multiple times: you count this when you choose $1$ and $2$ in the binomial coefficient $\binom{7}{2}$ and from the remaining $5$ goods $\{3,4,5,6,7\}$, good $3$ goes to bidder $1$; but also when you choose $1$ and $3$ in the binomial coefficient $\binom{7}{2}$ and from the remaining $5$ goods $\{2,4,5,6,7\}$, good $2$ goes to bidder $1$, etc. Of course these distributions are the same and should not be counted multiple times.
I hope this clears it up for you.