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There are $N$ boxes, which are marked as $1,2, \ldots, N$. We want to put $n$ balls into these boxes. If we require that each box has $1$ ball for $n$ selected boxes, what's the probability? Of course, $n\leq N$. There are two ways to solve it.

1. We make some wings on these balls, and let them fly into the boxes.

The first ball has $N$ cases, the same as the second and so on. Thus, there are $N^n$ cases in all.

As for our requirement, there are $n$ balls fly into the selected $n$ boxes.

That is $n!$ cases in all.

Hence, the answer is $$\frac{n!}{N^n}$$

Of course, there is another way another way.

2. These boxes offer $N+1$ positions for these balls. Let these balls park in them, there are $C_{N+1}^{n}$ cases.

Then we add some sticks over the empty positions, that will form $N$ boxes.

As for our requirement, the case which is satisfied, is only $1$ case.

Hence, the answer is $$\frac{1}{C_{N+1}^{n}}$$

Obviously, both answers are reasonable, but the results are not the same. As for the reason, I think solution 1 thinks that the balls are different, but the solution 2 doesn't think so.

So, which one is the correct solution? I am really confused...

N. F. Taussig
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Alexander Lau
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  • I'm not from the English speaking country, so there maybe some grammer mistakes. My apologize, but I'm happy if you could point out these language mistakes. My appreciate~ – Alexander Lau Nov 04 '18 at 09:08
  • Or in briefly speaking ,put 2 balls into 3 boxes. There are 9 cases or 6 cases? – Alexander Lau Nov 04 '18 at 09:22
  • Are the balls distinct? – N. F. Taussig Nov 04 '18 at 09:31
  • @N.F.Taussig The point is at here. I cut the problem from the book, but it doesn't tell us whether they are distinct. – Alexander Lau Nov 04 '18 at 09:39
  • The first is realistic while the second less so. Sadly books are not realistic. (I think perhaps your second example should be using $N+n-1$ rather than $N+1$) – Henry Nov 04 '18 at 10:07
  • @Henry Those $\binom{N + n - 1}{N - 1}$ cases are not equally likely. There is only one way to place all $n$ balls in one box, while there are $n!$ orders in which $n$ balls can be placed in $n$ different boxes. – N. F. Taussig Nov 04 '18 at 10:12
  • @N.F.Taussig - they are equally likely when the book says they are. That is why I suggested that "sadly books are not realistic". I do not think we disagree. https://math.stackexchange.com/questions/2975845/why-is-a-simulation-of-a-probability-experiment-off-by-a-factor-of-10 raises a similar issue – Henry Nov 04 '18 at 11:13

1 Answers1

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There are $N$ boxes into which each ball could be placed, so our sample space has $N^n$ elements.

If the balls are distinct, then there are $\binom{N}{n}$ ways to select the boxes in which the $n$ balls will be placed and $n!$ ways to distribute them to the selected boxes so that one ball will be placed in each box. Hence, the number of favorable cases is $$\binom{N}{n}n!$$ Hence, the probability that the $n$ balls will be placed in $n$ different boxes is $$\frac{\binom{N}{n}n!}{N^n}$$ For your example of placing two balls in three boxes, suppose the boxes are numbered from $1$ to $3$ and that we wish to place a blue ball and a green ball in the boxes. There are $3$ ways to place the blue ball and three ways to place the green ball, so there are $3^2 = 9$ ways to place the balls. As for the favorable cases, for each of the three ways we could place the blue ball, there are two ways to place the green ball, so there are $3 \cdot 2 = 6$ favorable cases. Alternatively, we could choose two of the three boxes in $\binom{3}{2}$ ways and distribute the two balls to the selected boxes so that there is one ball in each box in $2!$ ways. Notice that $\binom{3}{2}2! = 6$.

If the balls are indistinguishable, there are still $N^n$ elements in the sample space. To see why, take $n$ balls of different colors. Place them in the boxes. Then paint all the balls the same color to make them indistinguishable.

However, now the number of favorable cases is $\binom{N}{n}$ since it no longer matters which ball is placed in which box.

Hence, if the balls are indistinguishable, the probability that each ball is placed in a separate box is $$\frac{\binom{N}{n}}{N^n}$$ For your example of placing two balls in three boxes, suppose the boxes are numbered from $1$ to $3$ and that two indistinguishable red balls are placed in the boxes. There are still nine possible outcomes since there are three choices for the first ball we place and three choices for the second ball we place (alternatively, place a blue ball, then a green ball, then paint them both red after they have been placed). However, there are only three favorable outcomes, depending on which box is left empty or, alternatively, which two of the three boxes we choose to fill with a red ball. Notice that $\binom{3}{2} = 3$.

N. F. Taussig
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