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Basket ball players

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by j_shreyans » Sat Aug 30, 2014 8:52 pm
9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?

A)23
B)30
C)42
D)60
E)126

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by [email protected] » Sat Aug 30, 2014 10:28 pm
Hi j_shreyans,

This question does NOT ask us to put players "in order", it asks us for groups of players. That clue points to using the Combination Formula. This question has 2 types of players though (guards and forwards), so we have to use the Combination Formula twice (once for each type of player), then multiply the results.

Guards:
There are 6 guards and we're asked for sets of 3.

6c3 = 6!/(3!3!) = 6(5)(4)(3)(2)(1)/3(2)(1)(3)(2)(1) = 20 different sets of 3 guards

Forwards:
There are 3 forwards and we're asked for groups of 2.

3c2 = 3!/(2!1!) = 3(2)(1)/(2)(1)(1) = 3 different groups of 2 forwards

(20)(3) = 60 possible teams

Final Answer: D

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by Brent@GMATPrepNow » Sun Aug 31, 2014 7:01 am
j_shreyans wrote:9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?

A)23
B)30
C)42
D)60
E)126
Take the task of creating a team and break it into stages.

Stage 1: Select 3 guards from the 6 eligible guards
Since the order in which we select the guards does not matter, we can use combinations.
We can select 3 guards from the 6 eligible guards in 6C3 ways (= 20 ways)
So, we can complete stage 1 in 20 ways

If anyone is interested, we have a free video on calculating combinations (like 6C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Select 2 forwards from the 3 eligible forwards
Since the order in which we select the forwards does not matter, we can use combinations.
We can select 2 forwards from the 3 eligible forwards in 3C2 ways (= 3 ways)
So, we can complete stage 2 in 3 ways

By the Fundamental Counting Principle (FCP), we can complete the two stages (and thus create a basketball team) in (20)(3) ways ([spoiler]= 60 ways[/spoiler])

Answer: D
--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

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Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by GMATinsight » Tue Sep 02, 2014 9:00 pm
j_shreyans wrote:9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?

A)23
B)30
C)42
D)60
E)126

OAD
No. of ways to select 3 Guards out of 6 guards = 6C3 = 20
No. of ways to select 2 forwards out of 3 forwards = 3C2 = 3

Total possible ways = 20 x 3 = 60

Answer: Option D
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by Jeff@TargetTestPrep » Mon Mar 12, 2018 3:41 pm
j_shreyans wrote:9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?

A)23
B)30
C)42
D)60
E)126
The guards can be chosen in the following number of ways:

6C3 = 6!/[(3!(6 - 3)!] = 6!/)3!3!) = (6 x 5 x 4)/3! = (6 x 5 x 4)/(3 x 2 x 1) = 20 ways

The forwards can be selected in the following number of ways:

3C2 = (3 x 2)/2! = 3 ways

We can pair up each of the 20 ways of choosing guards with each of the 3 ways of choosing forwards. So the total number of teams of 3 guards and 2 forwards is 20 x 3 = 60.

Answer: D

Jeffrey Miller
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