Problem Solving

There are four contestants in the competition for 6 different prizes. Each contestant can win one prize. How many different outcomes are possible at the competition for the four contestants?

A) 15

B) 30

C) 60

D) 180

E) 360

OA is E. Should not it be A??

The problem is 6C4 = 6!/4!2! = 5*6/2 = 15

The actual explanation is:
This is a combination problem that uses factorials. You need to choose a combination of k=4 choices out of n=6 total choices, and use the formula: n!/(n-k)!

But n!/(n-k)! is a formula for permutations, not combinations.

Thanks!

There are 4 people in line to win 6 prizes.

_ _ _ _

the 1st person has 6 prize options he can claim from. (6)
the 2nd person has now only 5 prize options he can claim from. (6*5)
the 3rd person has 4 prize options he can claim from. (6*5*4)
the 4th person has 3 prize options he claim from. (6*5*4*3)

So it is 6*5*4*3 = 360 different outcomes are possible. (E)

oks you are right. i mean your method
but you forgot that those 4 prizes can be arranged in 4! different ways among the participants
so answer will be 15*4!
=360

tohellandback, how do you know when to use the combinations and when - permutations formula? since this particular problem doesn't say anything about the order/arrangements (hence, permutations)...

oks wrote:tohellandback, how do you know when to use the combinations and when - permutations formula? since this particular problem doesn't say anything about the order/arrangements (hence, permutations)...

Oks ,
tell you what I really don't know the difference. All I know is a few formulas and when to apply them..familiarize with the formulas ...take about 20 questions ..do them without looking at the explanations..think about them..
and after all that go through the explanations and do them again..I am sure you will know when to apply those formulas..

I used to be confused too..I am now much better that I used to be.

well you have to get to the very definitions of permutations and combinations to resolve the difference and when to use what. Combination problems usually apply to choosing out of available options "How many ways can you chose two girls from 40 girls" This is choosing a certain sample something out of a total set. Hence you use the C formula. Permutations is the different ways you can arrange/order things. Like two girls need to be seated in four chairs. Here you need to apply the P formula. Permutation problems tend to revolve around two discrete sets of data.

Here is a funny problem. try it. A speed dating event has ten girls and six guys. How many different ways can the pairs be formed (assume straight relationships)

The solution is simple. You can approach it multiple ways. One way is:
6P4. Another way is