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If six different balls are placed in three different boxes, what is the probability that exactly two balls are placed in the first box?

My attempt: $k_1+k_2+k_3=6$, $k_i$ - quantity of balls in $i$ box. $k_1=2$ hence there are 4 sets for $k_2, k_3:$ $\{0,4\},\{1,3\},\{3,1\},\{4,0\}$.
Adding set $\{2,2\}$ we get all combinations when there are 2 balls in the first box $\Rightarrow$ possibility $= \frac{4}{5}$?

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  • $\begingroup$ Are you asking what the probability is that two balls are placed in the first box if six balls are distributed to three boxes? $\endgroup$ – N. F. Taussig Sep 30 '17 at 0:37
  • $\begingroup$ yes, I do. English isn't my native language, hence sometime my states are bad to uderstand $\endgroup$ – ioleg19029700 Sep 30 '17 at 0:39
  • $\begingroup$ yes. Let's say they have numbers from 1 to 6 $\endgroup$ – ioleg19029700 Sep 30 '17 at 0:41
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There are $3^6$ ways to distribute six balls to three boxes since there are three choices for each of the six balls.

There are $\binom{6}{2}$ ways to select exactly two of the six balls to be placed in the first box and $2$ ways to select the box in which each of the four remaining balls is placed. Hence, there are $$\binom{6}{2}2^4$$ favorable cases.

Therefore, the probability that exactly two balls are placed in the first box when six balls are distributed to three different boxes is $$\frac{\binom{6}{2}2^4}{3^6}$$

Note: You attempted to count cases by considering how many balls are placed in each box. However, the $$\binom{6 + 3 - 1}{3 - 1} = \binom{8}{2}$$ solutions to the equation $$k_1 + k_2 + k_3 = 6$$ in the nonnegative integers are not equally likely to occur. There is only one way to place all six balls in the first box. However, there are $$\binom{6}{2}\binom{4}{2}\binom{2}{2} = 90$$ ways to place two balls in each box.

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  • $\begingroup$ +1 nice answerTaussig as usual...I did this at the numerator $\binom{6}{2}(\binom{4}{0}+\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4})=2^4\binom{6}{2}$ $\endgroup$ – Isham Sep 30 '17 at 1:33
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    $\begingroup$ @Isham That works. The number of subsets of a set with $n$ elements is $2^n$ since each element is either included or not included in a subset. Hence, a set with four elements has $2^4$ subsets. The number of subsets of size $k$ of a set with $n$ elements is $\binom{n}{k}$. Since the number of elements in a subset of a four-element set ranges from $0$ to $4$, $$\binom{4}{0} + \binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 2^4$$ $\endgroup$ – N. F. Taussig Sep 30 '17 at 1:39
  • $\begingroup$ Yes it was easier for me that way...Then I used Newton's binomial.. $\endgroup$ – Isham Sep 30 '17 at 1:40

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