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9 basketball players are trying out to be on a newly formed
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Updated on: 27 Jul 2014, 23:11
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Question Stats:
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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?
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21 Aug 2010, 15:53
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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
I hope it helps let me know if it does if it does not you can always PM me.....!!!! and don't forget to give a Kudos if you like the explanation.... _________________
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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
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.
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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!
Re: 9 basketball players are trying out to be on a newly formed
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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
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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.
Answer: D
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Re: 9 basketball players are trying out to be on a newly formed
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21 Apr 2018, 06:23
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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)
Answer: D
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
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21 Apr 2018, 06:23