business school case competition

BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

This topic has expert replies

Post new topic Post Reply

Previous Topic Next Topic

jantony345: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Sat Nov 03, 2012 6:14 pm

business school case competition

by jantony345 » Wed Apr 19, 2017 7:28 pm

In a business school case competition, the top three teams receive cash prizes of $5,000, $3,000, and $2,000, respectively, while the remaining teams are not ranked and do not receive any prizes. If there are six participating teams, how many outcomes of the competition are possible?

(A) 18
(B) 20
(C) 36
(D) 60
(E) 120

Source Veritas, OA: E

I picked 120 because I started out with 6P3. However, I doubted my answer was right because of the option to not receive prizes. I curious how others would walk through the logic of this problem.

Quote

Source: Beat The GMAT — Problem Solving | Wed Apr 19, 2017 7:28 pm

GMAT/MBA Expert

[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pm; Location: Palo Alto, CA; Thanked: 2867 times; Followed by:511 members; GMAT Score:800

by [email protected] » Wed Apr 19, 2017 8:25 pm

Hi jantony345,

The prompt states that the three remaining teams (the ones who do NOT win a prize) are NOT ranked, so the order of those three teams is NOT a factor in the calculation. In addition, 6! = 720 - and that result is not among the answer choices, so it clearly cannot be the answer.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

Quote

GMAT/MBA Expert

Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 am; Location: Grand Central / New York; Thanked: 470 times; Followed by:34 members

by Jay@ManhattanReview » Wed Apr 19, 2017 9:03 pm

jantony345 wrote:In a business school case competition, the top three teams receive cash prizes of $5,000, $3,000, and $2,000, respectively, while the remaining teams are not ranked and do not receive any prizes. If there are six participating teams, how many outcomes of the competition are possible?

(A) 18
(B) 20
(C) 36
(D) 60
(E) 120

Source Veritas, OA: E

I picked 120 because I started out with 6P3. However, I doubted my answer was right because of the option to not receive prizes. I curious how others would walk through the logic of this problem.

Hi jantony345,

Any of the six teams can win any one of the three prizes.

# of ways the first prize can be won = 6;

# of ways the second prize can be won = 6 - 1 = 5;

# of ways the third prize can be won = 5 - 1 = 4

Since it does not matter which teams rank fourth, fifth and sixth as there is no prize for those positions.

Total # of ways = 6*5*4 = 120.

Total # of ways = 6P3 = 120 is also the correct way. Since the order matters, for example, Team A winning the first prize and Team B winning the second prize is different from Team B winning the first prize and Team A winning the second prize.

The correct answer: E

Hope this helps!

Relevant book: Manhattan Review GMAT Combinatorics and Probability Guide

-Jay
_________________
Manhattan Review GMAT Prep

Locations: New York | Singapore | London | Dubai | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

Quote

GMAT/MBA Expert

Scott@TargetTestPrep: GMAT Instructor; Posts: 8085; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Fri Apr 28, 2017 2:03 pm

jantony345 wrote:In a business school case competition, the top three teams receive cash prizes of $5,000, $3,000, and $2,000, respectively, while the remaining teams are not ranked and do not receive any prizes. If there are six participating teams, how many outcomes of the competition are possible?

(A) 18
(B) 20
(C) 36
(D) 60
(E) 120

We need to determine in how many ways 6 teams can be ranked first, second, or third in a competition. Thus, in this case, order matters, and so we have a permutation problem. We need to know in how many ways we can arrange 6 teams for 3 places. Thus:

6P3 = 6!/3! = 6 x 5 x 4 = 120 ways.

Answer: E

Scott Woodbury-Stewart
Founder and CEO
[email protected]