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In a business school case competition, the top three teams receive cas 31 Oct 2015, 07:32
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85% (hard)

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55% (02:37) correct 45% (02:45) wrong based on 391 sessions

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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. There are 6 participating teams, named Team A, Team B, Team C, Team D, Team E, and Team F. If Team A wins one of the prizes, Team B will also win one of the prizes. How many outcomes of the competition are possible?

A. 18
B. 20
C. 54
D. 84
E. 120
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D
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In a business school case competition, the top three teams receive cas 31 Oct 2015, 07:44
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shasadou
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. There are 6 participating teams, named Team A, Team B, Team C, Team D, Team E, and Team F. If Team A wins one of the prizes, Team B will also win one of the prizes. How many outcomes of the competition are possible?

A. 18
B. 20
C. 54
D. 84
E. 120
Show more


the ways the three teams can be selected is..
i) A is selected, B is also selected, third can be any of the remaining 4.. so total ways 4.. they can be arranged in 4*3!=24 ways..
ii) A is not selected. threee teams can be selected from remaining 5 so total arrangements =5C3*3!=60
total =84 outcomes
D
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In a business school case competition, the top three teams receive cas 24 Sep 2018, 23:01
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This may be approached in the following two ways:

Quote:
If there weren’t any constraints at all, this problem would have a simple solution: Six teams are competing for three slots, so N = 6 and K = 3. Plugging in these values into our equation for a permutation we discover:

A=N!/(N−K)!=6!/(6−3)!=6!/(3)!=6∗5!/6=120

Because the answer must be less than 120, we know the answer is not “E”.

Calculating the number of options without Team A is also easy. There are five remaining teams (B through F) vying for three possible slots (N = 5, K = 3). Because order matters, we plug these values into our equation for permutations:

A=N!/(N−K)!=5!/(5−3)!=5!/(2)!=120/2=60

Because 60 doesn’t even take into account the possibility of Team A winning, we know that the actual number of arrangements will be above 60, so answer choices “A”, “B”, and “C” cannot be correct. There is only one option between 60 and 120 in the answer choices: 84.

The answer is “D”.
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Quote:
Alternately, we could calculate the number of arrangements where Team A is in the winner’s circle. Every time this happens, Team B will be there somewhere. Sketching this out might be the easiest to see. Here are the possible combinations:

AB? A?B ?AB BA? B?A ?BA

Notice that this is nothing more than a permutation of three elements: A, B, and ?, where “?” represents another team. There are 3! = 6 possible arrangements. This is also similar to a subgroup problem – for every arrangement of A, B, and ?, we can fit in four possible teams into the “?” slot. Therefore, the number of possible arrangements where A is included is 6 × 4 = 24. We could add these arrangements to the possibilities where Team A isn’t in the winner’s circle (60) to arrive at a total: 60 + 24 = 84. The answer is still “D”.

We could also take the total without constraints (120) and subtract the number of arrangements that wouldn’t work (where A is in the winner’s circle, but B isn’t.)
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Notice there are multiple ways of solving many high-level combinatorics questions: by excluding answers using easy logic, by calculating pieces and adding them together, or by assuming there aren’t any constraints and subtracting the exceptions that wouldn’t work.
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In a business school case competition, the top three teams receive cas 05 Oct 2018, 01:29
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why are they considering the option of A not winning the prize when its already given that A wins a prize B also wins..we have to select one team out of 4..
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In a business school case competition, the top three teams receive cas 06 Oct 2018, 02:49
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One of the quicker and easier approach:-

Three ranks are there to occupy.
Six teams are contenders.

Two possible cases (add them):-

1st Case: If 'A' is NOT selected:-
Three ranks are up for grab among 5 teams. So, it can be occupied in \(5 * 4 * 3 = 60\) ways.


2nd Case: If 'A' is selected:-
B would also get selected. So among 3 ranks,
2 ranks can be occupied by A & B in \(3P2\) = \(6\) ways
1 rank can be occupied by either of the other 4 teams in \(4\) ways.
\(6 * 4 = 24\) ways


Add both cases, to get all possible ways:-
\(60 + 24 = 84\) (answer)



shasadou
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. There are 6 participating teams, named Team A, Team B, Team C, Team D, Team E, and Team F. If Team A wins one of the prizes, Team B will also win one of the prizes. How many outcomes of the competition are possible?

A. 18
B. 20
C. 54
D. 84
E. 120
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In a business school case competition, the top three teams receive cas 09 Oct 2018, 10:41
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shasadou
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. There are 6 participating teams, named Team A, Team B, Team C, Team D, Team E, and Team F. If Team A wins one of the prizes, Team B will also win one of the prizes. How many outcomes of the competition are possible?

A. 18
B. 20
C. 54
D. 84
E. 120
Show more

We have two cases to consider: 1) A is one of the top three teams, and 2) A is not one of the top three teams.

Case 1: A is one of the top three teams

If A is one of the top three teams, then B is also one of the top three teams. We only have 4C1 = 4 ways to choose the third top team. In other words, we have 4 possible sets of top three teams (or winning teams). However, for each set of 3 winning teams, there are 3! = 6 ways for how they win the prizes. Therefore, there are 4 x 6 = 24 possible outcomes of the competition if A is one of the top three teams.

Case 2: A is not one of the top three teams

If A is not one of the top three teams, we could have 5C3 = 10 ways to choose the top three teams. (Note that Team B could be one of the top three teams, with Team A NOT being in the top three.) In other words, we have 10 possible sets of top three teams (or winning teams). Similar to case 1, for each set of 3 winning teams, there are 3! = 6 ways for how they win the prizes. Therefore, there are 10 x 6 = 60 possible outcomes of the competition if A is not one of the top three teams.

Therefore, there are a total of 24 + 60 = 84 possible outcomes of the competition.

Answer: D
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In a business school case competition, the top three teams receive cas 17 Feb 2025, 04:44
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The Trap is that A can win any of the prizes, it does not say that B will win AFTER A has won, so in this case A can win the 3rd prize as well. We have 3 spots:

_ _ _

Let's take the case where A wins:

A can take any of the 3 spots (including the 3rd) = 3

If A has won, B must win too, B can take any of the 2 remaining spots: 2

The last spot can be taken by any of the 4 remaining players: 4

Number of combinations in which A is a winner = 3*2*4 = 24

Now, let us count the number of cases in which A does not win. Notice here that B will surely win if A does but B's win is not contingent on A winning, that is, it is not a necessary condition for B to win, he can win without A winning but if A does win, B will win too, if A does not win, B may or may not win.

Again, we have 3 spots:

_ _ _

The first spot can be taken by any player except A = 5

The second spot can be taken by any of the remaining players = 4

The third spot can be taken by any of the remaining players = 3

Total = 5*4*3 = 60

Total cases = Cases where A wins + Cases where A does not win = 24 + 60 = 84
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In a business school case competition, the top three teams receive cas 28 Mar 2026, 06:12
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