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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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13 Oct 2009, 06:28
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81% (01:00) correct 19% (01:01) wrong based on 200 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
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Jul 2014, 00:11, 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, 10:07
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Agree on 60 , 6C3 * 3C2
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13 Oct 2009, 10:15
6C3 x 3C2
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18 Aug 2010, 13:17
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, 02:59
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, 16:53
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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

I hope it helps let me know if it does if it does not you can always PM me.....!!!!
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25 Feb 2011, 08:54
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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

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, 09:04
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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, 09:28
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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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27 Jul 2014, 21:30
Hello from the GMAT Club BumpBot!

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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, 05:10
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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26 Feb 2016, 08:50
Hello from the GMAT Club BumpBot!

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Re: 9 basketball players are trying out to be on a newly formed   [#permalink] 26 Feb 2016, 08:50
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