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Daily Magic Spell Discussion - January 12, 2017

 
 
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Daily Magic Spell Discussion - January 12, 2017
InstituteAreteem 发表于 2017年02月9日 Thursday 13:23
 

Here is another question that I noticed that most users got incorrect. 

Suppose a football team has $15$ members. The team plays two games, and must choose a captain and co-captain for both games. Suppose that the captain for the first game cannot be a captain or co-captain for the second game (but the first co-captain can be either captain again). How many ways can the team choose a captain and co-captain for the two games?

The official solution is provided below:

For the first game, we will select a captain and then select a co-captain. Since there are $15$ members on the team, there are $15$ options for the captain. After the captain is selected, there are $14$ options for the co-captain. This implies that for the first game, there are $15 \times 14  = 210$ possibilities to select a captain and co-captain for the first game.

For the second game, we will again select a captain and then select a co-captain. The captain from the first game is not allowed to be the captain or the co-captain, so this implies that there are $15-1 = 14$ members to choose for the second game. Using the similar method discussed above, there are $14 \times 13 = 182$ possibilities to select a captain and co-captain for the second game.

Therefore, there are $210 \times 182 = 38220$ possible ways of choosing a captain and a co-captain for the two games.

I believe that again, the most common mistake is misinterpretation of the problem. For this particular problem, organizing who is captain and co-captain can be quite tricky since both roles are leadership positions. Because of this, students are inclined to confuse themselves on what they are counting, yielding different answers to the problem.

Some students may naively assume that both captain and co-captain are two different people so they count the number of ways to choose $2$ people from a group of $15$. This is done in $\binom{15}{2} = 105$ ways. After the two people are chosen per game, students might be inclined to think that there are $105$ options for the first game and $105$ options for the second game, yielding a total of $105^2 = 11025$ ways. Note that this method assumes that both captain and co-captains are different people. The only unsatisfied condition is that the first co-captain can be either captain again.

Let me know if you had any other approaches that sounds reasonable to you! Please post below if you have any other thoughts about this problem.