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

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Manager
Joined: 12 Feb 2012
Posts: 125
There are 3 teams each with 5 basket players. How many combi  [#permalink]

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01 Sep 2013, 13:16
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Difficulty:

25% (medium)

Question Stats:

80% (02:12) correct 20% (01:52) wrong based on 49 sessions

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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?
Intern
Joined: 09 Jun 2013
Posts: 49
GMAT 1: 680 Q49 V33
GMAT 2: 690 Q49 V34
GPA: 3.86

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01 Sep 2013, 22:27
1
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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Intern
Joined: 09 Feb 2017
Posts: 9
Re: There are 3 teams each with 5 basket players. How many combi  [#permalink]

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28 Apr 2017, 07:50
Can someone explain to me why the correct solution is 3C2x5C1x5C1? I am confused as to how we decided to use that equation. When I attempted the problem, I thought the equation would be 5C2x5C2x5C2
Intern
Joined: 13 Mar 2016
Posts: 12
Location: Singapore
Concentration: Technology, Operations
WE: Information Technology (Consulting)
Re: There are 3 teams each with 5 basket players. How many combi  [#permalink]

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28 Apr 2017, 08:12
1
Team A- 5 players ; Team B-5 players ; Team C- 5 players
To select 2 players who are not from the same team.
Hence possiblities ..to select from Team A &B - (5C1)×(5C1)= 25
OR (+)
Team B & C - (5C1)×(5C1)= 25
OR(+)
Team C & A - (5C1)×(5C1)= 25

Total Combinations= 25+25+25=75
Re: There are 3 teams each with 5 basket players. How many combi &nbs [#permalink] 28 Apr 2017, 08:12
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