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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.2012-03-19

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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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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 $1-2/3=1/3$, but I don't know how she or he came up with that complicated expression for that probability.

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    can you try and see the problem in my way? the 3C2 is for choosing a couple of boys, the 1C1 * 7C1 is for chossing the leftout boy a matching girl, then 6C2 * 4C2 * 2C2 is for choosing girl couples, and finally, divide by 5! to disregard the order...2012-03-19
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    @Ofek: I would have tried to see the problem your way right away if you had tried to explain the way you saw it :-) Now that you explained it, I see what you were trying to do and where you went wrong: The factor $5!$ for the order of the couples is too much because you didn't actually count each of $5!$ orders once -- you treated the couple of boys and the mixed couple separately without counting different orders, so you only have to divide by $3!$ for the order of the three couples of girls. That yields exactly the factor of $20$ by which your result was off.2012-03-19
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    @joriki Could you please explain to me in more detail why deviding by $5!$ is too much? I did a calculation very similar to that of Ofek namely: $\frac{{3\choose2} {8\choose2}{6\choose2} {4\choose2} {2\choose 2} / 5!}{{10\choose2} {8\choose2} {6\choose 2} {4\choose 2} {2\choose 2} / 5!}$, but I seem to be off by a factor 5. My guess is that I should only devide by $4!$ in the numerator as this gives the desired result, but could you please explain why this is so?2013-07-05
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    I mean, aren't there still 5 different couples for which we multiplied the binomial coefficients with eachother (hence introducing order between the couples each time)?2013-07-05
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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.2013-07-06
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    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.2013-07-06
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    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.2013-07-06
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    @joriki Thank you very much for your help and explanation. The only thing that I do not fully understand yet, is how you can deduce the following: "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". How could I see this myself? The same for 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"-how did you see that?2013-07-06
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    @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.2013-07-06
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    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.2013-07-06
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    @joriki Ah, that's exactly the explanation that I needed! Now it makes a lot of sense :). Thank you very much for your help :)!2013-07-06