No, it isn't.
I think the answer is 108 ... but I'm not exactly sure.
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?
5 answers
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.
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.
Your thinking is absolutely right. I made the mistake of counting only 18 ways for 2nd and 3rd (18*6 = 108).
Sry.
Sry.
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
that makes it 6*5*4 possibilities=120 different combinations for 1st,2nd and 3rd
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 permutations and not combinations.
Permutations imply positioning, while combinations do not.
So there are 6*5*4 or 120 ways, as Jordan said.
Except Jordan meant to say
120 different permutations and not combinations.
Permutations imply positioning, while combinations do not.