SOLUTION: Kindly answer this question please!!! Thanks!!! A gold, a silver, and a bronze medal are to be awarded to any 3 of the 12 contestant. In how many

Algebra ->  Permutations -> SOLUTION: Kindly answer this question please!!! Thanks!!! A gold, a silver, and a bronze medal are to be awarded to any 3 of the 12 contestant. In how many       Log On


   

Question 217312: Kindly answer this question please!!! Thanks!!! A gold, a silver, and a bronze medal are to be awarded to any 3 of the 12 contestant. In how many ways can the medals be given if A.anyone can be a winner B.The gold medal is to be awarded to a particular person?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
12 contestants.
if anyone can be a winner, the number of possible combinations is given by the formula %28n%29%21%2F%28n-x%29%21 which becomes 12%21%2F%289%21%29+=+12%2A11%2A10+=+1320
-----
Assuming the winner is chosen, then we are talking about 2 slots left out of 11 contestants which then becomes 11%21%2F%289%21%29+=+11%2A10+=+110
-----
To make this easier to show, assume the number of contestants is 4.
3 slots for the winners would be 4%21%2F%281%21%29+=+4%2A3%2A2+=+24 possible ways.
Let the contestants be a,b,c, and d.
The possible ways for the winners would be:
abc
abd
acb
acd
adb
adc
bac
bad
bca
bcd
bda
bdc
cab
cad
cba
cbd
cda
cdb
dab
dac
dba
dbc
dca
dcb
-----
Now assuming d was chosen to be first.
The remaining slots were to be filed with a, b, or c.
-----
the formula would be %283%21%29%2F%281%21%29+=+3%2A2+=+6 different ways the second and third positions would be filled.
Those would be:
dab
dac
dba
dbc
dca
dcb
-----