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Six highschool boys gather at the gym for a modified game [#permalink]

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20 Oct 2013, 13:45

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Six highschool boys gather at the gym for a modified game of basketball involving three teams. Three teams of 2 people each will be created. How many ways are there to create these 3 teams?

Six highschool boys gather at the gym for a modified game of basketball involving three teams. Three teams of 2 people each will be created. How many ways are there to create these 3 teams?

(A) 15 (B) 30 (C) 42 (D) 90 (E) 108

\(\frac{C^2_6*C^2_4*C^2_2}{3!}=15\) (dividing by 3! because the order of the teams doesn't matter).

Re: Six highschool boys gather at the gym for a modified game [#permalink]

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20 Oct 2013, 20:31

Bunuel wrote:

stunn3r wrote:

Six highschool boys gather at the gym for a modified game of basketball involving three teams. Three teams of 2 people each will be created. How many ways are there to create these 3 teams?

(A) 15 (B) 30 (C) 42 (D) 90 (E) 108

\(\frac{C^2_6*C^2_4*C^2_2}{3!}=15\) (dividing by 3! because the order of the teams doesn't matter).

Re: Six highschool boys gather at the gym for a modified game [#permalink]

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21 Oct 2013, 11:43

AccipiterQ wrote:

Bunuel wrote:

stunn3r wrote:

Six highschool boys gather at the gym for a modified game of basketball involving three teams. Three teams of 2 people each will be created. How many ways are there to create these 3 teams?

(A) 15 (B) 30 (C) 42 (D) 90 (E) 108

\(\frac{C^2_6*C^2_4*C^2_2}{3!}=15\) (dividing by 3! because the order of the teams doesn't matter).

Why does team order not mattering result in the 3! ending up in the divisor?

Hi,

It is because we are just asked to select 3 teams and not concerned with relative order. Say we have selected 3 teams ABC , now the other orders BCA or CAB are the same teams. total there are 6 outcomes if we consider the orders

I think for this purpose we can just have 6C2 teams , meaning we are just finding 2 items out of six to form a team which yield 15
_________________

Re: Six highschool boys gather at the gym for a modified game [#permalink]

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21 Oct 2013, 12:06

stunn3r wrote:

Six highschool boys gather at the gym for a modified game of basketball involving three teams. Three teams of 2 people each will be created. How many ways are there to create these 3 teams?

(A) 15 (B) 30 (C) 42 (D) 90 (E) 108

To put it in a different way: Say 6 boys are a,b,c,d,e,f and we need to form 3 teams say t1,t2,t3

With a : a,b a,c a,d a,e a,f --->5 With b : b,c b,d b,e b,f --->4 With c : c,d c,e c,f --->3 With d : d,e d,f --->2 With e : e,f --->1 With f : --- --->0

Re: Six highschool boys gather at the gym for a modified game [#permalink]

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22 Oct 2013, 09:27

Bunuel wrote:

stunn3r wrote:

Six highschool boys gather at the gym for a modified game of basketball involving three teams. Three teams of 2 people each will be created. How many ways are there to create these 3 teams?

(A) 15 (B) 30 (C) 42 (D) 90 (E) 108

\(\frac{C^2_6*C^2_4*C^2_2}{3!}=15\) (dividing by 3! because the order of the teams doesn't matter).

I do not understand why there is no division for each team by 2? For a given team, it does not matter if player a was select first and then player b, or visa versa. Pls. help

Re: Six highschool boys gather at the gym for a modified game [#permalink]

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22 Oct 2013, 09:35

ronr34 wrote:

Bunuel wrote:

stunn3r wrote:

Six highschool boys gather at the gym for a modified game of basketball involving three teams. Three teams of 2 people each will be created. How many ways are there to create these 3 teams?

(A) 15 (B) 30 (C) 42 (D) 90 (E) 108

\(\frac{C^2_6*C^2_4*C^2_2}{3!}=15\) (dividing by 3! because the order of the teams doesn't matter).

I do not understand why there is no division for each team by 2? For a given team, it does not matter if player a was select first and then player b, or visa versa. Pls. help

Hi Ron,

Your question is not clear. Can you please elaborate.

Re: Six highschool boys gather at the gym for a modified game [#permalink]

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22 Oct 2013, 09:36

Bunuel wrote:

stunn3r wrote:

Six highschool boys gather at the gym for a modified game of basketball involving three teams. Three teams of 2 people each will be created. How many ways are there to create these 3 teams?

(A) 15 (B) 30 (C) 42 (D) 90 (E) 108

\(\frac{C^2_6*C^2_4*C^2_2}{3!}=15\) (dividing by 3! because the order of the teams doesn't matter).

Answer: A.

Hope it helps.

Bunuel,

I solved it differently: \(C^2_6\) to get the number of ways for one team to be created multiplied by 3 for three teams. = \(C^2_6\) * 3 =15 Is this logic flawed?
_________________

Re: Six highschool boys gather at the gym for a modified game [#permalink]

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22 Oct 2013, 09:46

kulkarnios wrote:

ronr34 wrote:

I do not understand why there is no division for each team by 2? For a given team, it does not matter if player a was select first and then player b, or visa versa. Pls. help

Hi Ron,

Your question is not clear. Can you please elaborate.

The numerator is all the options to take all the teams (choose two from 6, then choose 2 from 4) The we divide this by 3! so that we take into account all possible permutations of the teams. But for each team, of player a and player b, it is the same team if a is chosen first and then b, just as if b were chosen first and then a. So to my logic, we need to divide each team for example 6C2 by 2 to take that into account.... no?

lets says for the 1st team i chose any two boys ( so it is 6c2 i.e 15 ways ) now two members are out and i am left with 4 boys ,,,now i agn chose 2 boys out of 4 ( 4c2....6 ways )

now only two members will be left and they will form the third team

so total no of ways 15 * 6 = 90

You have selected the "first team", the "second team" and the "third team" i.e. you have ordered the teams. Say, you select A and B for first team, C and D for second team and E and F for the third team. Now say you select C and D for first team, A and B for second team and E and F for third team. Are they different selections? No. We just need 3 teams. The three teams are A and B, C and D and E and F in both cases. So we need to divide the teams we obtain by 3! to get rid of the team arrangement.

Re: Six highschool boys gather at the gym for a modified game [#permalink]

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30 Sep 2014, 06:36

VeritasPrepKarishma wrote:

Quote:

where am i going wrong in this :

lets say we have 1,2,3,4,5,6 boys

we need to make 3 teams of two members each

lets says for the 1st team i chose any two boys ( so it is 6c2 i.e 15 ways ) now two members are out and i am left with 4 boys ,,,now i agn chose 2 boys out of 4 ( 4c2....6 ways )

now only two members will be left and they will form the third team

so total no of ways 15 * 6 = 90

You have selected the "first team", the "second team" and the "third team" i.e. you have ordered the teams. Say, you select A and B for first team, C and D for second team and E and F for the third team. Now say you select C and D for first team, A and B for second team and E and F for third team. Are they different selections? No. We just need 3 teams. The three teams are A and B, C and D and E and F in both cases. So we need to divide the teams we obtain by 3! to get rid of the team arrangement.

90/3! = 15 (correct answer)

it is always there if we look....thanks for the explanation and taking the time out .... +1

Re: Six highschool boys gather at the gym for a modified game [#permalink]

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01 Apr 2016, 01:09

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