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

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
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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