Prize Combo

[tttrn333]

In a business school case competition, the top three teams receive cash prizes of $5,000,$3000,$2000.
If there are 6 participating team, A,B,C,D,E,F. If A wins one prize and B wins another, how many outcomes are possible?

How I am solving:

Since A,B need to be in and since order matters, I assume its a set of 3. A,B, and O (other). Ways to arrange them is 3!=6

Since there are 4 different ways O can be (C,D,E,F). Answ= 6*4=24.

This is however wrong. Can someone explain?

[amit2k9]

A,B can win prizes in 3c2 * 2! ways = 6
3c2 - 3 positions among which any 2 can be filled. 2! - between AB the combinations possible.

last position can be taken by 4c1 ways
thus, when AB are present = 6* 4 = 24 ways

when A is not present, remaining positions can be filled by any of the 5 teams
hence 5*4*3 = 60 ways

thus total = 60+24 = 84 ways.

[Anurag@Gurome]

In a business school case competition, the top three teams receive cash prizes of $5,000,$3000,$2000.
If there are 6 participating team, A,B,C,D,E,F. If A wins one prize and B wins another, how many outcomes are possible?

How I am solving:

Since A,B need to be in and since order matters, I assume its a set of 3. A,B, and O (other). Ways to arrange them is 3!=6

Since there are 4 different ways O can be (C,D,E,F). Answ= 6*4=24.

This is however wrong. Can someone explain?

The last part of the problem should read as "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?"

Solution:
This means that Case 1) If A is selected, then B is also selected. So, the total number of all possible arrangements is 4C1 * 3! = 24

Case 2) If A is not there, anybody can be selected. Note that the question does not mean that 'If A is not there, B will not be there.' Here the number of arrangements possible is 5P3 = 60.

So, total number of outcomes possible is 24+60 = 84.

[tttrn333]

ok i see how you guys are interpreting it. is it me or is this question ambiguous?