# Six men and six women are paired to dance. What is the probability each pair is wearing the same color shirt?

Six men and six women are paired to dance. Two men wear red shirts, two men wear white shirts, and two men wear blue shirts. Similarly, two women wear red shirts, two women wear white shirts, and two women wear blue shirts. The men and women are paired off into couples (one man, one woman) at random. What is the probability all the couples are wearing the same color shirts?

My answer is $\frac{1}{90}$. There are $6! = 720$ ways to pair the couples and for each color there are $2$ ways to pair the men and women such that partners have the same color shirt. Thus there are $2^3=8$ pairings in which all partners wear the same color shirt, and $\frac{8}{720} = \frac{1}{90}$. But my textbook says the answer is $\frac{1}{9}$.

• I suppose you're assuming that every dancing couple consists of one man and one woman. – David K Aug 14 '17 at 2:07
• Yes, that's implied in the problem. Math has yet to get woke. – Colin Weinshenker Aug 14 '17 at 2:12
• 6-factorial is not 36. – Gerry Myerson Aug 14 '17 at 2:37
• @ColinWeinshenker Are you sure the textbook doesn't give the answer as $\frac{1}{90}$ – WW1 Aug 14 '17 at 2:41
• Yeah, I think it might've been a typo. – Colin Weinshenker Aug 14 '17 at 2:45

Your claim that the probability is $\frac 8{6!}$ is correct, but $6! \neq 36$. $\frac 8{6!}=\frac 8{720}=\frac 1{90}$ If you line up the women with the red shirts, then the white shirts, then the blue, the chance that the men's shirts match is $\frac 13 \cdot \frac 15 \cdot \frac 12 \cdot \frac 13\cdot 1 \cdot 1=\frac 1{90}$
There are $6!$ distinct and equally probable ways to arrange the women, of which $2!^3$ will match colours with the men, so the probability of this happening is indeed $8/720$ or $1/9\color{silver}0$.
Opps, the ink faded on that last zero.   It is supposed to be $1/90$.