Six highschool boys gather at the gym for a modified game

Question

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

Bunuel · Answer

\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. For more on this check: a-group-of-8-friends-want-to-play-doubles-tennis-how-many-55369.html in-how-many-different-ways-can-a-group-of-9-people-be-85993.html probability-88685.html probability-85993.html combination-55369.html sub-committee-86346.html Hope it helps.

AccipiterQ · Answer

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

adg142000 · Answer

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

kulkarnios · Answer

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

Hence Total is 5+4+3+2+1+0 = 15.

Hence 15 such teams are possible

-- Omkar

In with with 'kudos' -- gmatclub

KarishmaB · Answer

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)

sahilvijay · Answer

LET ME HAVE PLEASURE TO INTRODUCE A SMALL METHOD

6 PERSONS
3 TEAMS
2 PERSON IN EACH TEAM

NO OF WAYS = 6!/(3!2!2!2!)
that is = 2! will come the n number of times here as n=3 so they are coming 3 times 2!2!2!

6!/3!2!2!2! = 15

Ans is A

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Six highschool boys gather at the gym for a modified game

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Six highschool boys gather at the gym for a modified game

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards