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This is studying for my final, not homework.

Six balls are dropped at random into ten boxes. What is the probability that no box will contain two or more balls?

So I know that to select 6 balls from ten we do: ${10\choose 6}$, but I'm not sure how to check whether or not each box contains less than two balls. Is it just:

${6\choose 1}\over{10\choose 6}$

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There are $10^6$ different ways in which the six balls could land in the ten boxes. Now we'll count the number of ways that have each ball in a different box. If each ball is in a different box, they must occupy $6$ of the $10$ boxes. There are $\binom{10}6$ different sets of six boxes that they could occupy, and they can be permuted in $6!$ ways amongst those boxes, so there are altogether $$\binom{10}6\cdot 6!=\frac{10!}{4!}$$ ways in which they can all fall into different boxes. The desired probability is therefore $$\frac{\frac{10!}{4!}}{10^6}=\frac{10!}{4!\cdot10^6}=\frac{9!}{4!\cdot10^5}=0.1512\;.$$

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Drop a ball into a box. Now drop the next ball. There are $9$ empty boxes, so the probability of no collision is $\frac{9}{10}$. Then drop the next one. The probability of no collision is $\frac{8}{10}$. And so on.

So the overall probability of no collision is $$\frac{9}{10}\cdot\frac{8}{10}\cdot\frac{7}{10}\cdot\frac{6}{10}\cdot\frac{5}{10}.$$

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