Questão 02: (2 pontos) Durante suas férias, oito amigos, dos quais dois são canhotos, decidem realizar um torneio de vôlei de praia. Eles precisam formar quatro duplas para a realização do torneio. Nenhuma dupla pode ser formada por dois jogadores canhotos. De quantas maneiras diferentes podem ser formadas essas quatro duplas?

<p> This question involves combinatorics, a topic within mathematics that deals with counting, arrangement, and probability. To solve this problem, we consider the different ways to form pairs for the volleyball tournament under the given condition: no pair can have two left-handed players.<br /><br />We have 8 friends, of which 2 are left-handed. We need to form 4 pairs for the tournament.<br /><br />1. First, we pair the two left-handed players with right-handed players. There are 6 right-handed players, so there are 6 choices for the first left-handed player and 5 choices for the second. This results in 6 * 5 = 30 possible pairings for the left-handed players.<br /><br />2. Now, we need to pair the remaining 4 right-handed players. These can be paired in a standard way of pairing n people into n/2 pairs, which is given by the formula (n-1)!! (double factorial). For 4 people, it's (4-1)!! = 3!! = 3 * 1 = 3.<br /><br />3. The total number of ways to form the pairs is the product of these two steps. So, it's 30 * 3 = 90.<br /><br />Thus, there are 90 different ways to form the four pairs under the given conditions.</p>