5. Six people (two wearing red shirts, two wearing blue shirts, two wearing yellow shirts) want to divide themselves into three teams of two people each to play briscola. In how many ways can they divide themselves so that the two in each team have shirts of different colors?
(A) 24
(B) 8
(C) 11
(D) 4
(E) 15
Problem 5. The answer is (B).

The two people with red shirts cannot be on the same team; furthermore, their teammates must have shirts of different colors, otherwise, in the remaining team, there would end up being two people with shirts of the same color. One red player is therefore paired with a blue player, while the other red player is paired with a yellow player. The composition of these two pairs also determines the third. We have two possible choices for the red player who plays with a blue one; the other red player will team up with a yellow one. At this point, we have two choices for which blue player plays with the first red and which yellow player plays with the second red. All choices are independent, and we therefore have a total of $2 \cdot 2 \cdot 2=8$ ways to pair the players.

(Problem proposed by A. Bianchi)