business school

akhp77 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 and prizes. There are 6 participating teams, names 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?

There are 3 prize locations to be filled from 6 contestants.

The given constraint is unidirectional, “if A then B” is not logically same as “if B then A”. Hence B could win a prize money without finding A collecting another.

If A wins a prize, two prizes are booked, and the remaining single prize could go to one of the remaining 4 contestants. Never forget that the three prizes could be swapped among the chosen three in 3! manners again. This could total to 1 × 1 × 4 × 3! = 24 ways.

These are the appearances of B when A was there too. Now, let’s analyse the case when A didn’t win a prize; now, B could or couldn’t win a prize; also remember, there’s no A now, so even if B is there among the prizes, this is not repeating a case with A there among the prizes.

If A doesn’t win a prize, the 3 prize locations could now be filled from 5 contestants in 5P3 = 60 ways.

[spoiler]24 + 60 = 84[/spoiler] many outcomes of the competition are possible.

akhp77 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 and prizes. There are 6 participating teams, names 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?

at a high level, there are 2 possibilities 1. A wins a prize and 2. A does not win a prize

Incase , A wins a prize ..that means B also wins a prize .. So the 3rd team which wins a prize among the 4 remaining teams can be selected in 4 ways... Now the 3 teams can be arranged in 3! ways to arrange the 3 teams
So total number of ways = 3! * 4 = 24 ways

Incase , A does not win a prize
3 teams can be selected from 5 teams to win prizes = 5C3 (note that B can still win a prize )
again 3! ways to arranges the teams to the prizes
So total number of ways = 5C3 * 3! = 10 * 6 = 60

So total number of outcomes = 84

i agree with @deb and 84 is correct

above explanation is fine but I could not understand why 5P3 is needed. I believe that I was struggling on this only.

The condition is =
If team A wins one of the prizes, Team B will also win one of the prizes.

Why we need to consider "A does not win a prize"?

akhp77 wrote:above explanation is fine but I could not understand why 5P3 is needed. I believe that I was struggling on this only.

The condition is =
If team A wins one of the prizes, Team B will also win one of the prizes.

Why we need to consider "A does not win a prize"?

Why do you support A so much, anything under-table with A?

I believe I understood it now.

It is not mandatory that A is always there. I was confused initially.

Thanks