Question

Did I answer this problem right?
There are 6 contestants in a singing competition. How many different ways can first, second and third place be awarded?
Here my answer: 6![(6-3)!*3]=4*5*6/2*3=20
Is this correct?

Answer 1

No, it isn't.

I think the answer is 108 ... but I'm not exactly sure.

Answer 2

I don't think so.

There are people who can get first
six people who can get second
six peoplewho can get third.

Let the six folks be a, b, c, d, e, f

abc
abd
abe
abf
acb
acd
ace
acf
adb
adc
ade
adf
aeb
aec
aed
aef
afb
afc
afd
afe
so if a gets first, there twenty ways to get second and third.

So by the same logic, b,c,d,e,f can be first, so
ways: 20*6=120

check my thinking.

Answer 3

Your thinking is absolutely right. I made the mistake of counting only 18 ways for 2nd and 3rd (18*6 = 108).
Sry.

Answer 4

an easier way to think about it is that you have 6 ways to choose first place and considering that the same person can not get second place you only have 5 to choose from for second place and of course 4 people to choose from for third place.
that makes it 6*5*4 possibilities=120 different combinations for 1st,2nd and 3rd

Answer 5

There are 6 ways to fill first, then 5 ways to fill second, and 4 ways to hand out third prize.
So there are 6*5*4 or 120 ways, as Jordan said.

Except Jordan meant to say
120 different <b>permutations</b> and not combinations.

Permutations imply positioning, while combinations do not.