# I've come up with two solutions to this problem, and I don't know which is correct.

If I have 5 distinct pair pairs of gloves, 10 distinct gloves in all, how many ways are there to distribute two gloves to each of 5 children if the two gloves someone receives can also be two left- or right-handed gloves? I'm assuming 10 distinct gloves is because of the left and right hand fit.

Soltn 1:

Suppose I threw all 10 distinct gloves in a bag. Then suppose each of the 5 kids stood in a line and stuck their hands out. If I were to go down the line, picking a glove from the bag per hand, I would have 10*9*...*2*1 = 10! ways to distribute the gloves. Since there aren't any identical gloves, I wouldn't be dividing by anything.

Soltn 2:

Suppose again they all stood in a line. For each kid, I took 2 out of the bag. So for the first kid, $\binom{10}{2}$, for the second kid, $\binom{8}{2}$, and so on. But the product of all those combinations is less than 10! by a factor of $2^{-5}$

What accounts for this discrepancy?

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Label the gloves $1$ through $10$. The first solution overlooks the fact that handing them out in the order $1,2,\color{red}{3,4},\color{blue}{5,6},\color{green}{7,8},\color{magenta}{9,10}$ and handing them out in the order $2,1,\color{red}{3,4},\color{blue}{6,5},\color{green}{7,8},\color{magenta}{10,9}$, for instance, give exactly the same gloves to the same kids. You can interchange the first and second gloves given out, and it makes no difference to the kids. Similarly, you can interchange the third and fourth, or the fifth and sixth, or the seventh and eighth, or the ninth and tenth. That’s five pairs that can go either way, so you have to divide by $2^5$ to get rid of the overcounting.