Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In a business school case competition, the top three teams [#permalink]
05 Jul 2012, 22:02

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

41% (03:24) correct
59% (02:03) wrong based on 132 sessions

In a business school case competition, the top three teams receive cash prizes of $1000, $ 2000, and $ 3000. while remaining teams are not ranked and do not receive any prizes, there are 6 participating teams, named A, B, C, D, E, F. If team A wins one of the prizes, team B wins also one of the prizes. How many outcomes of the competition are possible ?

Re: In a business school case competition, the top three teams [#permalink]
06 Jul 2012, 00:14

2

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

alchemist009 wrote:

In a business school case competition, the top three teams receive cash prizes of $1000, $ 2000, and $ 3000. while remaining teams are not ranked and do not receive any prizes, there are 6 participating teams, named A, B, C, D, E, F. If team A wins one of the prizes, team B wins also one of the prizes. How many outcomes of the competition are possible ?

A. 18 B. 28 C. 36 D. 84 E. 120

We are told that "if team A wins one of the prizes, team B wins also one of the prizes". Consider following cases:

A wins one of the prizes, then B must also win one of the prizes, and in this case we can have 4 triplets: {ABC}, {ABD}, {ABE}, {ABF}. Each triplet can be arranged in 3!=6 ways. Hence in the case when A wins one of the prizes 4*6=24 arrangements are possible.

A does NOT win one of the prizes, then three winners must be from other 5 teams. 3 winners out of 5 (B, C, D, E, F) teams can be chosen in \(C^3_5=10\) ways and each case (for example {CDE}) can be arranged in 3!=6 ways, hence in the case when A does NOT win one of the prizes 10*6=60 arrangements are possible.

Re: In a business school case competition, the top three teams [#permalink]
12 Apr 2014, 20:33

Option D. If we make cases: I:\(A\) doesn't win:\(5*4*3=60\) cases + II:\(A\) wins:\(C(3,1)*C(2,1)*C(4,1)=24\) cases Because \(A\) could take any one of three prizes \(B\) could take any of the 2 prizes left And the third leftover prize could be taken by any one of \(C,D,E,F\). Total=\(84\) cases

Re: In a business school case competition, the top three teams [#permalink]
13 Apr 2014, 01:55

1

This post received KUDOS

Bunuel wrote:

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

Alternative solution:

If there would have been no constraint the number of possible scenarios were: 6P\(3\) = 6!/3! = 120 Now we need to calculate cases where A is a winner but B is not So the other two positions can be taken by C, D , E, F . total number of combination = 4C2 = 4!/(2!*2!) = 6 As the three winners can be arranged among themselves in 3! ways, total number of outcomes with A as winner but no B = 6*6 = 36

Hence the number of outcomes which satisfy the constraint in the question = 120 - 36 = 84.

------------------------------------------ Kudos if the answer helped

Re: In a business school case competition, the top three teams [#permalink]
14 Apr 2014, 00:47

ind23 wrote:

Bunuel wrote:

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

Alternative solution:

If there would have been no constraint the number of possible scenarios were: 6P\(3\) = 6!/3! = 120 Now we need to calculate cases where A is a winner but B is not So the other two positions can be taken by C, D , E, F . total number of combination = 4C2 = 4!/(2!*2!) = 6 As the three winners can be arranged among themselves in 3! ways, total number of outcomes with A as winner but no B = 6*6 = 36

Hence the number of outcomes which satisfy the constraint in the question = 120 - 36 = 84.

------------------------------------------ Kudos if the answer helped

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...