# Given that $6$ men and $6$ women are divided into pairs, what is the probability that none of the women will sit with a man?

I've generalized the question I was given here for simplicity: $6$ men and $6$ women are to be paired for a bus trip. If the pairings are done randomly, what's the probability that no women will end up sitting next to a man? Here's my first attempt, but I'm really not sure whether this is the right way to find the desired probability.

We want to group $12$ people into $6$ single-sex groups of $2$, so I began by calculating the number of ways we can make $6$ pairs. By the multinomial function: $12!/(2!)^6=7484400$ ways to make $6$ pairs.

Then, we want to figure out the number of ways we can make single-sex pairs. Again, by the multinomial function, we have $6!/(2!)^3=90$ ways to make $3$ female-female pairs. Since we also have to consider male pairs, I squared this to get $8100$ ways to make $6$ single-sex pairs.

I found that the likelihood of all pairs being single-sex is $8100/7484400=0.00108$, but this doesn't seem like a completely reasonable probability.

Could you help me find the errors in my method for solving this problem?

• It looks right to me. It might just be a rare to happen. Feb 11, 2016 at 6:50
• Argh! I thought you said $\displaystyle{6! \over (2!)^3\cdot 3!}=5\cdot3\cdot 1$. So yes, your numbers are wrong. (Also, $\displaystyle{12!\over (2!)^6 \cdot 6!}$ is the correct number of ways to pair up the people.) Feb 11, 2016 at 7:05
• So then, for single-sex pairs, you would find that there are $6!/(2!)^3 3!=15$ possible ways to make female-female pairs, which makes 225 total ways to make all single-sex pairs--is that what you're implying? Feb 11, 2016 at 7:12
• The probability is $(5/11)(3/9)(1/7)$. Minimal counting. Feb 11, 2016 at 7:15
• @cembos1005 Yes. See my comment after David's answer. Feb 11, 2016 at 7:18

We want to group $12$ people into $6$ single-sex groups of $2$, so I began by calculating the number of ways we can make $6$ pairs.

By the multinomial function: $12!/(2!)^6=7484400$ ways to make $6$ pairs.

No, order doesn't matter within the groups of $2$ nor of the $6$ groups.

$$\frac{12!}{2!^6 6!} = 10395$$

Then, we want to figure out the number of ways we can make single-sex pairs. Again, by the multinomial function, we have $6!/(2!)^3=90$ ways to make $3$ female-female pairs. Since we also have to consider male pairs, I squared this to get $8100$ ways to make $6$ single-sex pairs.

$$\left(\frac{6!}{2!^3 3!}\right)^2 = 225$$

I found that the likelihood of all pairs being single-sex is $8100/7484400=0.00108$, but this doesn't seem like a completely reasonable probability.

$$\frac{6!^3}{12!3!^2}= \frac{5}{231}$$

This is also equal to $\frac{5}{11}\frac{3}{9}\frac{1}{7}$, the probability that a girl is paired with a girl, another girl is paired with a girl, and that the last two girls are paired.

There is an error here.

In the general case, you considered pairings to be different if the order of the six pairs was different.

In the single-sex case, there is an ordering, but only within each of the two groups of three pairs.

To correct the error, it would probably be easiest to consider selections of six pairs without taking into account the way the six pairs are ordered.

Edit Here is how I would solve the problem in the quickest way. Let's first pick a mate for the first man. What is the probability this will be a man? It's $5/11$. Assume that occurs. The probability the third man will be paired with a man is $3/9$. Finally, the likelihood the last two men will be matched is $1/7$. So the answer is $(5/11) \times (1/3) \times (1/7) = 5/231 \approx 0.0216$.

To stay as close as possible to your method, you need to divide your general answer by the number of ways to order the six pairs, which is $6!$. For the second part, divide each time by $3!$. So the answer will be $$\frac{(6!/[(2!)^3 3!])^2}{12!/[(2!)^6 6!]}$$

• No there isn't. If you arbitrarily sort the women into A, B, C, D, E, F, you can ask: How many ways are there to match up the first unmatched woman in the list (in this case, A)? Answer: 5. How many ways are there to match the first unmatched woman in the list (whether it's B or C)? Answer: 3, and now there's only one pair left. ((ADDENDUM:)) The method is right; I misread what the OP calculated this answer as. Feb 11, 2016 at 7:02
• What would be the best way to do this without considering order? Obviously this is a Combinations problem without replacement, but I thought the multinomial function took care of the order issue, as it's of the n C m form. Feb 11, 2016 at 7:05
• Dividing by $2!$ several times took care of the ordering within the pairs. However, you failed to take into account the ordering of the six (or three) pairs themselves. Feb 11, 2016 at 7:14
• @CarlHeckman I don't understand exactly what your objection is to what I wrote. Feb 11, 2016 at 7:21
• @David You said there was an error. I originally said, no there isn't, then re-read the OP, and you were right; there was an error. (I thought the OP had included an extra $6!$ or $3!$ in getting his numbers.) Feb 11, 2016 at 7:23

When in doubt about your solution to a certain problem, you should always try to approach the problem differently and see if you get the same answer. If the answers differ, figuring out where the difference lies is itself an interesting exercise.

E.g. in this case compute the chance of the first pair being a single-sex pair, then compute the chance of the second pair being a single-sex pair, etc..

1st pair being single-sex:

$$\frac{5}{11}$$

2nd pair being single-sex given 1st pair is single-sex:

$$\frac{6}{10}\cdot\frac{5}{9} + \frac{4}{10}\cdot\frac{3}{9}$$

3rd pair being single-sex given 1st & 2nd pair are single-sex:

$$(\frac{6}{8}\cdot\frac{5}{7} + \frac{2}{8}\cdot\frac{1}{7})+(\frac{5}{8}\cdot\frac{4}{7} + \frac{3}{8}\cdot\frac{2}{7})+(\frac{4}{8}\cdot\frac{3}{7} + \frac{5}{8}\cdot\frac{4}{7})+(\frac{3}{8}\cdot\frac{2}{7} + \frac{6}{8}\cdot\frac{5}{7})$$

....

after that multiply them together to obtain the chance of having 6 single-sex pair, and simply simplify the multiplicative expression to see if it is equivalent to the expression you initially came up with.

And now you may notice that there is no point keep tracking each pair. All you need is to compute the chance of 3 pairs being single-sex of the same sex, and the remaining pairs of the opposite sex would also be single-sex.

With so many approaches already posted, you might like the conceptually simple $\dfrac{\binom63}{\binom{12}{6}}$, although it lacks the minimalist elegance of $\frac5{11}\frac39\frac17$