# Probability Combinatorial related; choosing couples

7 girs and 3 boys are divided to couples, order within a couple and between couples is not important, what is the probability that one of the couples contains 2 boys?

i had this exercise in my final exem and im not realy sure, am i wrong?

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Imagine yourself in that situation -- can you imagine that you'd get a couple of boys only $1$ out of $60$ times? Even just the probability that the very first couple drawn has two boys, $(3/10)(2/10)=6/100$, is quite a bit bigger than that. –  joriki Mar 19 '12 at 13:06

Imagine $10$ slots in $5$ pairs to put the children in. To have all three boys in different couples, you have $10$ options to place the first boy, then $8$ to place the second, then $6$ to place the third. Then you can place the girls in the remaining slots any way you like, which gives another $7!$ possibilities. On the other hand, without restrictions, there are $10!$ possibilities. Thus, if you randomly pick some assignment with uniform distribution over all possible assignments, the probability of having all three boys in different couples is

$$\frac{10\cdot8\cdot6\cdot7!}{10!}=\frac{8\cdot6}{8\cdot9}=\frac23\;.$$

Thus your professor is right that the chance of getting two boys in one couple is $1-1/3=2/3$, but I don't know how she or he came up with that complicated expression for that probability.

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@rbm: If I understand correctly what you're doing, you seem to be making a similar mistake as Ofek. You only have to divide through for duplicates that occur in different orders if they actually do so occur. In Ofek's calculation, the boys couple could only occur as first choice, the mixed couple could only occur as second choice, and only the three girls couples could appear as any of the remaining three choices, yielding a factor $3!$ for the orders in which they can occur. –  joriki Jul 6 '13 at 13:05
In your calculation, the boys couple can again only occur as the first choice, but then the mixed couple and the girls couples can all occur as any of the remaining four choices, yielding a factor $4!$ for the orders in which they can occur. –  joriki Jul 6 '13 at 13:07
Imagine a slot for each of the five binomial coefficients you write down. Each slot can be filled with a pair. The first slot, corresponding to $\binom32$, is always filled with two boys, whereas if you had all $5!$ different permutations of the pairs, this slot would sometimes have to contain other pairs, and the boys pair would sometimes have to occur in a different slot. –  joriki Jul 6 '13 at 13:09
@rbm: I'm not sure how to explain further. Your first choice is a choice of two boys. The boys couple can only be chosen in this choice; it cannot be chosen in any other choice. By contrast, when you then make four choices of $2$ out of $8$, the mixed couple, for instance, can be chosen in any of those four choices. You're counting as different pairings e.g. $(B_1,B_2)(B_3,G_1)(G_2,G_3)(G_4,G_5),(G_6,G_7)$ and $(B_1,B_2)(G_2,G_3)(G_4,G_5)(G_6,G_7)(B_3,G_1)$, and there are $4!$ such equivalent pairings, so you have to divide by $4!$ to correct for that. –  joriki Jul 6 '13 at 13:47
By contrast, you are not counting e.g. $(B_3,G_1)(G_2,G_3)(G_4,G_5),(G_6,G_7)(B_1,B_2)$; this doesn't occur because you always choose the boys couple as the first pair. The factor $5!$ would only be correct if all $5!$ permutations of each pairing would occur in the way you choose pairs, but they don't. –  joriki Jul 6 '13 at 13:48

We describe correct reasoning that leads to the professor's form of the answer.

First let us count the number of ways to divide the people into couples. Line up the children in order of student number, or weight, or any criterion you like. The first child in the list can choose her/his partner in $9$ ways. For each way that this is done, there are $8$ kids left over. The first child in this list of unchosen $8$ can choose her/his partner in $7$ ways. Once this is done, the first unchosen child can choose her partner in $5$ ways, and so on, for a total of $9\cdot7\cdot5\cdot 3\cdot 1=945$ ways. All these ways are equally likely. That's the denominator of your professor's answer. (The $945$ could have been obtained in several other ways, and might be quoted from the solution of a previous problem.)

Now we count the number of ways of coupling people such that there is a two boy couple. (There can be at most $1$ such couple.) If we were going to count this in a way which is kind of like we counted the denominator, we would say that there are $\binom{3}{2}$ ways to choose the two boy couple. Now we have $8$ people left over, and by the same kind of reasoning that led to the denominator, there are $7\cdot5\cdot 3\cdot 1$ ways to couple them, for a total of $\binom{3}{2}(7\cdot 5\cdot 3\cdot 1)$ ways. Divide. Almost everything cancels and we get $1/3$.

The professor counted the number of pairings with a two boy pairing somewhat differently. The two boy pair can be chosen in $\binom{3}{2}$ ways. Now we have $8$ people left over. The remaining boy's partner can be chosen in $7$ ways. Now we have $6$ girls left. There are $6!$ ways to line them up in a row. Divide them into couples in the obvious way, first girl in the list of $6$ paired with the second, and so on. But this overcounts by a factor of $3!$ because each of the couplings can occur in $3!$ different orders. And there is a further overcount factor of $2^3$, obtained by "switching order" within couples in all possible ways. That gives a total of $$\binom{3}{2}\cdot 7\cdot \frac{6!}{3!2^3},$$ almost exactly what was written down.

Remark: The wording of the problem is somewhat odd. The probability we are interested in does not depend on whether order within couples or between the couples is important. It is easier to count both total number and number of favourables using order, as in joriki's efficient solution.

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