Question 217312
12 contestants.
if anyone can be a winner, the number of possible combinations is given by the formula {{{(n)!/(n-x)!}}} which becomes {{{12!/(9!) = 12*11*10 = 1320}}}
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Assuming the winner is chosen, then we are talking about 2 slots left out of 11 contestants which then becomes {{{11!/(9!) = 11*10 = 110}}}
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To make this easier to show, assume the number of contestants is 4.
3 slots for the winners would be {{{4!/(1!) = 4*3*2 = 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
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Now assuming d was chosen to be first.
The remaining slots were to be filed with a, b, or c.
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the formula would be {{{(3!)/(1!) = 3*2 = 6}}} different ways the second and third positions would be filled.
Those would be:
dab
dac
dba
dbc
dca
dcb
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