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$18$ people are randomly divided into $3$ groups of size $5$, $6$ and $7$. Among the $18$ are John, Jack and James. What is the probability that:

1) What is the probability that both Jack and James are included in the 6-persons group?

2) What is the probability that John, Jack and James are each in a different group?

3) What is the probability that two of the three are in the same group and the third is in a different group?

Thanks in advance!

This is NOT a homework question. This was an exam question.

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1)16*15*14*13*ncr(12,5) – derivative Nov 19 '13 at 18:21
@Lost1 For the first $$\frac{{^6{C_2}}}{{^{18}{C_2}}} = \frac{5}{{51}}$$ – Ahmed Ali Nov 19 '13 at 18:33
@Lost1 for the second $$\frac{{\left( \begin{array}{l}5\\1\end{array} \right) \cdot \left( \begin{array}{l}6\\1\end{array} \right) \cdot \left( \begin{array}{l}7\\1\end{array} \right)}}{{\left( \begin{array}{l}18\\3\end{array} \right)}} = \frac{{35}}{{136}}$$ – Ahmed Ali Nov 19 '13 at 18:38
dude, write them in your question with your reasoning, then people can tell you if it is correct and let you know where you got it wrong – Lost1 Nov 19 '13 at 18:43
up vote 3 down vote accepted

Your answers (given in the comments) for the first two are correct (and rather cleverer than many correct solutions). In the first, $\binom{18}2$ is the number of pairs of individuals, and $\binom62$ is the number of pairs taken from the group of $6$. The fraction $$\frac{\binom62}{\binom{18}2}$$ is therefore the probability that any given pair of individuals will end up together in the group of $6$. In particular, it’s the probability that Jack and James both end up in this group.

Similarly, $\binom{18}3$ is the number of $3$-person subsets of the group of $18$, and $\binom51\binom61\binom71$ is the number that include one person from each of the three subgroups, so $$\frac{\binom51\binom61\binom71}{\binom{18}3}$$ is the probability that any given set of $3$ people will end up in different subgroups. In particular, it’s the probability that John, Jack, and James will end up in different subgroups.

You can use the same general approach for the third question. There are still $\binom{18}3$ sets of $3$ people, so that will be my denominator. In the numerator I want the number of sets of $3$ people that have two people in one group and one in another. There are $$\binom52\binom{13}1$$ that have two in the group of $5$ and one in another group; just fill out the rest of the numerator, and you’ll have the desired probability.

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Like this? $$\frac{{\left( \begin{array}{c}5\\2\end{array} \right) \cdot \left( \begin{array}{c}13\\1\end{array} \right) + \left( \begin{array}{c}6\\2\end{array} \right) \cdot \left( \begin{array}{c}12\\1\end{array} \right) + \left( \begin{array}{c}7\\2\end{array} \right) \cdot \left( \begin{array}{c}11\\1\end{array} \right)}}{{\left( \begin{array}{c}18\\3\end{array} \right)}}$$ – Ahmed Ali Nov 20 '13 at 5:17
@Ahmed: Yes, exactly. – Brian M. Scott Nov 20 '13 at 5:18
This was my answer on the exam. I missed up the second one by multiplying it by $3$, though. LOL. Thanks alot! – Ahmed Ali Nov 20 '13 at 5:21

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