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A party consisting of $5$ boys and $10$ girls is divided at random into five groups of three persons each.What is the chance that each group will have one boy?

Number of ways of dividing into 5 groups is $\frac{15!}{(3!)^55!}$,but then stuck.Please guide me to the answer.

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closed as off-topic by 6005, user223391, Claude Leibovici, Adam Hughes, Najib Idrissi Oct 1 '15 at 7:43

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  • 3
    $\begingroup$ Downvoter can you please tell me,what is wrong in this question? $\endgroup$ – diya Oct 1 '15 at 4:14
  • $\begingroup$ It seems to me there is nothing wrong with the question, the approach and progress towards a solution were clearly described. $\endgroup$ – André Nicolas Oct 1 '15 at 15:25
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I would rather not use the number of ways to divide into groups. It does not feel concrete enough.

Line up $15$ chairs in a row. There are $\binom{15}{5}$ equally likely ways to choose where the boys will sit.

There are $3^5$ ways to choose one chair in the first group of three, and one in the second group of three, and so on.

Another way: Line up the boys in order of student number. Now they in turn choose group members at random. The probability Boy 1 chooses two girls is $\frac{10}{14}\cdot \frac{9}{13}$. Given this has happened, the probability Boy 2 chooses two girls is $\frac{8}{11}\cdot \frac{7}{10}$.

Given these things have happened, the probability Boy 3 chooses two girls is $\frac{6}{8}\cdot \frac{5}{7}$. And now the probability Boy 4 chooses two girls is $\frac{4}{5}\cdot \frac{3}{4}$. Multiply.

A little playing with factorials will transform the above product into more compact form.

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Although unchivalrous,
it is simpler to just place the $5$ boys,
and let the girls fend for themselves !

The first boy has to be in some group.

The 2nd boy now has $14$ vacant slots, of which only $12$ are admissible,
similarly the 3rd boy has $13$ vacant slots of which only $9$ are admissible, and so on, thus

$Pr = \dfrac{12}{14}\cdot\dfrac{9}{13}\cdot\dfrac{6}{12}\cdot\dfrac{3}{11} = \dfrac{81}{1001}$

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