1
$\begingroup$

1.46. Seven balls are distributed randomly into seven cells. What is the probability that the number of cells containing exactly $3$ balls is $2$?

I am getting different answer from this solution manual. My argument is the following:

In order for us to have an arrangement where there are exactly $2$ cells with $3$ balls, we can follow the following procedure. We first decide which $3$ balls we want to put them together among all $7$ balls, and then decide to which cell we want to put them in. This gives us $\binom{7}{3} \binom{7}{1}$. Now, among the left $4$ balls, we choose $3$ balls to put them together, and choose one cell among the remaining $6$ empty cells. This gives us $\binom{4}{3} \binom{6}{1}$. Finally, we are left with one ball, and we have $5$ choices regarding where to put it. Putting them together, we have

$$P(X_3 = 2) = \frac{\binom{7}{3} \binom{7}{1} \binom{4}{3} \binom{6}{1} 5}{7^7}$$

But the solution manual says the answer should be

$$\frac{\binom{7}{2}\binom{7}{3}\binom{4}{3}5}{7^7}$$

Who is wrong and why?

$\endgroup$
  • $\begingroup$ I think, that the problem is, that you must divide by 2, because in your case you specify, that the first 3 goes to the one and the other 3 goes to the other cell. However, the order of cells doesn't matter. Therefore, divide by 2 and get it. $\endgroup$ – user13 Jan 21 '18 at 14:36
0
$\begingroup$

The manual is right. The difference between your solution and the books solution is a factor of $2.$ That is ${7\choose1}{6\choose1} = 2{7\choose 2}$

You have placed 3 balls in one of 7 cells and 3 balls in one of 6 remaining. But you have double counted at this step.

$\endgroup$
  • $\begingroup$ But I don't "see" who I am double counting... $\endgroup$ – 3x89g2 Sep 14 '17 at 19:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.