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There are 3 teams each with 5 basket players. How many combi

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There are 3 teams each with 5 basket players. How many combi [#permalink] New post 01 Sep 2013, 12:16
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There are 3 teams each with 5 basket players. How many combinations are there if we were to pick 2 players from the 3 teams such that no team was picked twice?

A) 50
B) 23
C) 75
D) 28
E) 45

I know the answer is (3C2)*(5C1)*(5C1)=75
What other methods could we employ to find the same answer??

One method I am trying but its not working is: So we have 3 teams 5 each, so 15 players. I label each player by number and by team. So teams A,B,C. and the players are 1_a, 2_a,3_a,4_a,5_a, 1_b, 2_b ... 5_b, 1_c, 2_c .. 5_c.

We have a two slots. 15*14=190 permutations. But some of them are duplicates and order doesn't matter (eg. {3_a,5_c}={5_c,3_a}). So we divide by the number of slots 2!. 190/2!=95. Now we also have another constraint "No two players can be chosen from the same team". How many many ways are two 2 players from the same team are chosen? From team A we have 5 players. 5*4 ways to move them around. But again order doesnt matter so we divide by the number of slots 2!. So 10 ways. Same thing for team B and C. So we have a total of 30 ways to pick two player from the same team. from the 95 I subtract 30 to get 65. What am I doing wrong?

Are there any other ways to solve this problem other than what I laid out here?
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Kudos [?]: 26 [0], given: 3

Re: There are 3 teams each with 5 basket players. [#permalink] New post 01 Sep 2013, 21:27
alphabeta1234 wrote:
There are 3 teams each with 5 basket players. How many combinations are there if we were to pick 2 players from the 3 teams such that no team was picked twice?


A) 50
B) 23
C) 75
D) 28
E) 45

I know the answer is (3C2)*(5C1)*(5C1)=75
What other methods could we employ to find the same answer??

One method I am trying but its not working is: So we have 3 teams 5 each, so 15 players. I label each player by number and by team. So teams A,B,C. and the players are 1_a, 2_a,3_a,4_a,5_a, 1_b, 2_b ... 5_b, 1_c, 2_c .. 5_c.

We have a two slots. 15*14=190 permutations. But some of them are duplicates and order doesn't matter (eg. {3_a,5_c}={5_c,3_a}). So we divide by the number of slots 2!. 190/2!=95. Now we also have another constraint "No two players can be chosen from the same team". How many many ways are two 2 players from the same team are chosen? From team A we have 5 players. 5*4 ways to move them around. But again order doesnt matter so we divide by the number of slots 2!. So 10 ways. Same thing for team B and C. So we have a total of 30 ways to pick two player from the same team. from the 95 I subtract 30 to get 65. What am I doing wrong?

Are there any other ways to solve this problem other than what I laid out here?


There are altogether three arrangements, which are we can select two members each from team A and B, each from team A and C, and each from team B and C. For each arrangement, there are altogether 25 possibilities (5*5) since each team has 5 players. So there are a total of 75 possibilities (25*3). I hope this method is easier for you to understand.
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Re: There are 3 teams each with 5 basket players.   [#permalink] 01 Sep 2013, 21:27
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