Problem 28. At a party, there are 12 children, including 3 pairs of twins. How many ways are there to distribute 6 blue hats and 6 red hats to the children, so that in each pair of twins, both children are wearing hats of the same color? So we can divide this problem into two cases. Either the case where all the twins have hats of the same color, or the case where that's not true. So case 1, all the twins have hats of the same color. Then we just have to decide whether all the twins have red hats or all the twins have blue hats. The other hats of the other color, there's 6 of them, and 6 children that aren't twins, so everyone who's not a twin has to have a hat of the other color. So there's only two possibilities, red hats for the twins or blue hats for the twins. Case 2 is more complicated. In case 2, we know that the three pairs of twins don't all have hats of the same color. This means that two pairs have hats of one color, and one pair has hats of the other color, since we know that each pair, both children in the pair have hats of the same color. So there's two ways to pick which hat color is the majority color for the twins. And then we have to pick one of the three pairs of twins to have hats of the other color. So this means there's six ways to assign hats to the twins. For each of the two ways to select the majority color, there's three options for who has the minority hat color. Now, after we've assigned the hats to the twins, we have to assign hats to everyone else in the party. So there's six children left, and there's two hats left of the majority color for the twins and four of the other color. So we have to pick two of the children that aren't twins to have the two hats that are of the color that's most common for the twins. So there's six ways to pick the first other child to have this hat color and five ways to pick another child to have this hat color. But it doesn't matter which order we pick the children in. If we pick child one and then child two, it's the same as picking child two and then child one. So we're over counting every pair by two. So we have to divide by two. So six times five divided by two means there's 15 ways to distribute the remaining hats. So in total, we had six ways to assign hats to the twins in this case and then 15 ways to assign hats to everyone else. So in case two, there's a total of 90 ways. Now across the two cases, there were two ways in case one and 90 ways in case two, which makes 92 ways. And the answer is C.

party

children

twins

hat distribution

color combinations