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Basketball Representatives

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by EOV » Mon Dec 03, 2012 10:23 pm
Hi everybody,

Given that there are 5 basketball players per team, how many ways can you select 2 basketball
players from 3 teams if no more than one player can be selected from each team?

(A) 15
(B) 30
(C) 60
(D) 75
(E) 90

Please, help me to decide this task.
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by Anindya Madhudor » Mon Dec 03, 2012 11:19 pm
Let's say we have team A, B, and C.

Select 1 member from team A and 1 member from team B. This can be done in 5C1 * 5C1 = 25 ways.

Similarly, you can select 1 member from team A and 1 member from team C. And also you can select 1 member from team C and 1 member from team B.

Total number of selection = 25 *3 =75
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by Brent@GMATPrepNow » Tue Dec 04, 2012 7:13 am
EOV wrote:Hi everybody,

Given that there are 5 basketball players per team, how many ways can you select 2 basketball
players from 3 teams if no more than one player can be selected from each team?

(A) 15
(B) 30
(C) 60
(D) 75
(E) 90
Take the task of selecting 2 players and break it into stages.

Stage 1: Select the 2 teams from which you will select 1 player each.
Since the order in which we select the 2 teams does not matter. We can use combinations.
There are 3 teams, and we must select 2 of them.
This can be accomplished in 3C2 ways (3 ways).

Stage 2: From one of the selected teams, select a player
There are 5 players on that team, so we can accomplish this stage in 5 ways.

Stage 3: From the other selected team, select a player
There are 5 players on that team, so we can accomplish this stage in 5 ways.

By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus select 2 players) in (3)(5)(5) ways ([spoiler]= 75 ways = D[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775

We also have a free video on calculating combinations (like 3C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
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by GMATGuruNY » Tue Dec 04, 2012 7:30 am
EOV wrote:Hi everybody,

Given that there are 5 basketball players per team, how many ways can you select 2 basketball
players from 3 teams if no more than one player can be selected from each team?

(A) 15
(B) 30
(C) 60
(D) 75
(E) 90

Please, help me to decide this task.
Number of options for the first player selected = 15. (Any of the 15 players.)
Number of options for the second player selected = 10. (Any of the 10 players not on the same team as the first player.)
To combine these options, we multiply:
15*10.
Since the ORDER of the selections doesn't matter -- selecting AB is the same as selecting BA -- the product above must be divided by the number of ways to ARRANGE the two players selected (2!):
(15*10)/(2*1) = 75.

The correct answer is D.
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by sonalibhangay » Wed Dec 05, 2012 7:29 am
Please tell me if this way is right.

Total Number of Selections - Ways to select 2 players from one team

=15C2 - 3 ( 5C2)

=105 - 30

[spoiler]=75[/spoiler]

[spoiler]ANS = 75[/spoiler]
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