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# 9 basketball players are trying out to be on a newly formed

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9 basketball players are trying out to be on a newly formed  [#permalink]

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Updated on: 27 Jul 2014, 23:11
5
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Difficulty:

5% (low)

Question Stats:

84% (01:29) correct 16% (01:55) wrong based on 290 sessions

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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

Originally posted by hogann on 13 Oct 2009, 05:28.
Last edited by Bunuel on 27 Jul 2014, 23:11, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: 9 basketball players are trying out to be on a newly formed  [#permalink]

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13 Oct 2009, 09:07
1
Agree on 60 , 6C3 * 3C2
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Re: 9 basketball players are trying out to be on a newly formed  [#permalink]

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18 Aug 2010, 12:17
How do you get 6C3 * 3C2 ... ?
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Re: 9 basketball players are trying out to be on a newly formed  [#permalink]

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19 Aug 2010, 01:59
Financier wrote:
How do you get 6C3 * 3C2 ... ?

Out of 6 gaurds we have to select 3 -> selection means we use C -> so 6C3

Out of 3 forwards we have to select 2 -> selection means we use C -> so 3C2

Total ways = 6C3 x 3C2 = 60

Hope this helps!
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Re: 9 basketball players are trying out to be on a newly formed  [#permalink]

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21 Aug 2010, 15:53
2
1
Ok answer is 6C3 * 3C2

lets solve this question as
You have 6 positions and you need to place 3 people on that positions
How many ways you can do that
6C3 ways

Similarly you have 3 positions and you want 2 people to take that position in how many ways they can do that
3C2 ways

and they are mutually exclusive events i.e. there is no dependency of selection of guards on selection of forwards and vice versa
hence they should multiply

Therefore ,it is 6C3 * 3C2

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25 Feb 2011, 07:54
3
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

Sol:
3 guards of 6 guards AND 2 forwards of 3 forwards

$$C^6_3*C^3_2$$
$$\frac{6!}{3!3!}*\frac{3!}{2!1!}$$
$$\frac{6*5*4}{3*2}*\frac{3*2}{2}=60$$

Ans: "d"
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25 Feb 2011, 08:04
1
D. 60.
Choose 3 guards out of 6 * choose 2 forwards out of 3.
(6!/(3!*3!)) * (3!/(2!*1!) = 20 *3 = 60.
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25 Feb 2011, 08:28
1
You have to proceed with 2 separate combinations:

6C3, which is the number of ways in which you can select 3 guards out of 6. This yields 20
3C2, which is the number of ways in which you can pick 2 forwards out of 3. This yields 3.

Multiply both and you get 60. Answer: D
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Re: 9 basketball players are trying out to be on a newly formed  [#permalink]

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12 Dec 2014, 04:10
Hi guys,

just wondering if anyone could help me out with how to know whether the order matters. At first i thought it was just 6*5*4 times 3*2. I've been through a course and even with that i find it difficult to know when to apply combinations or permutations. Anyone have clear advice on this? Thanks in advance!
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Re: 9 basketball players are trying out to be on a newly formed  [#permalink]

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08 Mar 2018, 20:09
Hi All,

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

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Re: 9 basketball players are trying out to be on a newly formed  [#permalink]

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12 Mar 2018, 15:34
hogann 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.

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Re: 9 basketball players are trying out to be on a newly formed  [#permalink]

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21 Apr 2018, 06:23
Top Contributor
hogann 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

ASIDE: We have a video on calculating combinations (like 6C3) in your head (see below)

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 (= 60 ways)

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Re: 9 basketball players are trying out to be on a newly formed   [#permalink] 21 Apr 2018, 06:23
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