Combi

[MBA.Aspirant]

The University of Maryland, University of Vermont, and Emory University have each 4 soccer players. If a team of 9 is to be formed with an equal number of players from each university, how many number of ways can the selections be done?

A) 3
B) 4
C) 12
D) 16
E) 25

My question is why can't we choose from the total? i.e 12C9 is it because they want an equal number from each team?

Thanks

[vikram4689]

My answer is not in options - 4*4*4 =64

[Ian Stewart]

I discussed this question on another forum recently in a thread with a few questions from the same source, so I'll cut and paste the relevant portions. Apparently the source thinks the answer ought to be 12 to this question, which they arrive at by adding the number of choices from each university (there are 4 choices from each, since there are 4 people you can choose to leave off of the team). As I replied on the other forum:

The wording of most of these questions is pretty hopeless, but no matter how I read the soccer question, I can't see any interpretation that would make addition the correct thing to do. We add in counting problems when we have different ways of arriving at the result we're interested in - that is, when there are several cases to consider. That's not what is happening in this question; here we have choices from different universities, and when we make a sequence of choices, we need to multiply.

Mind you, the question "how many numbers of ways can the selections be done" doesn't make any sense to begin with, so there isn't much point debating what it might mean. This is a pedantic point, but suppose we take that literally, and decipher what it might mean. If there are 100 people in a room, and I ask the question "how many numbers of people are there in the room?" the answer is one: there is one number of people in the room, and that number is equal 100. It's an almost nonsensical question to ask, and while it may be 'technically... grammatically correct', it's mathematically meaningless.

Were you to see this question on the GMAT, it would surely be set up in such a way that you would multiply your choices from each university, so that's the version of the question that might be worthwhile to study. Under the only interpretation I find reasonable here, the answer ought to be (4C3)(4C3)(4C3) = 4^3 = 64.

I'd add that you cannot just find 12C9 here, because that would count *all* of the ways of choosing 9 people from the 12 available, including those selections where 4 players come from, say, Emory University. You don't want to count a lot of those possibilities.

[MBA.Aspirant]

Thanks for your reply. Can you please explain why 4^3? Thanks

[vikram4689]

Each university has to choose 3 players out of 4 players. This can be done in 4C3 =4 ways. For all the 3 univ. to make 1 team, no. of ways = 4*4*4 = 64.

[MBA.Aspirant]

This approach makes more sense. Like if you're choosing 2 things out of 5 and 3 things out of 4, the total ways is 40 ways not 14.

When should you add the outcomes and when should you multiply them?

[krishnasty]

Why isn't 64 in the list????
What's the OA??

[vikram4689]

@MBA.Aspirant
When you achieve ONE result by combining different factors(as in this case) MULTIPLY the factors
When you achieve ONE result by each of the different factors ADD the factors

@krish: dont worry ans is 64

[MBA.Aspirant]

@MBA.Aspirant
When you achieve ONE result by combining different factors(as in this case) MULTIPLY the factors
When you achieve ONE result by each of the different factors ADD the factors

@krish: dont worry ans is 64

Thanks Vikram. Don't wanna sound greedy but examples?

[vikram4689]

Well i ll give a rather simple example. Lets say there is a 3 course meals having dishes (A,B,C)

1. In how many ways i can eat a dish = 3 (1+1+1) - i can eat any of the dishes.

2. In how many ways i can eat 3 course meal = 1 (1*1*1) - i need to eat all the dishes to complete the 3 course meal.

[amit2k9]

4c3*4c3*4c3 = 64 it is.