document.write( "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? \n" ); document.write( "

Algebra.Com's Answer #163914 by Theo(13342) $\"\"$ $\"About$

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12 contestants.
\n" ); document.write( "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 $\"12%21%2F%289%21%29+=+12%2A11%2A10+=+1320\"$
\n" ); document.write( "-----
\n" ); document.write( "Assuming the winner is chosen, then we are talking about 2 slots left out of 11 contestants which then becomes $\"11%21%2F%289%21%29+=+11%2A10+=+110\"$
\n" ); document.write( "-----
\n" ); document.write( "To make this easier to show, assume the number of contestants is 4.
\n" ); document.write( "3 slots for the winners would be $\"4%21%2F%281%21%29+=+4%2A3%2A2+=+24\"$ possible ways.
\n" ); document.write( "Let the contestants be a,b,c, and d.
\n" ); document.write( "The possible ways for the winners would be:
\n" ); document.write( "abc
\n" ); document.write( "abd
\n" ); document.write( "acb
\n" ); document.write( "acd
\n" ); document.write( "adb
\n" ); document.write( "adc
\n" ); document.write( "bac
\n" ); document.write( "bad
\n" ); document.write( "bca
\n" ); document.write( "bcd
\n" ); document.write( "bda
\n" ); document.write( "bdc
\n" ); document.write( "cab
\n" ); document.write( "cad
\n" ); document.write( "cba
\n" ); document.write( "cbd
\n" ); document.write( "cda
\n" ); document.write( "cdb
\n" ); document.write( "dab
\n" ); document.write( "dac
\n" ); document.write( "dba
\n" ); document.write( "dbc
\n" ); document.write( "dca
\n" ); document.write( "dcb
\n" ); document.write( "-----
\n" ); document.write( "Now assuming d was chosen to be first.
\n" ); document.write( "The remaining slots were to be filed with a, b, or c.
\n" ); document.write( "-----
\n" ); document.write( "the formula would be $\"%283%21%29%2F%281%21%29+=+3%2A2+=+6\"$ different ways the second and third positions would be filled.
\n" ); document.write( "Those would be:
\n" ); document.write( "dab
\n" ); document.write( "dac
\n" ); document.write( "dba
\n" ); document.write( "dbc
\n" ); document.write( "dca
\n" ); document.write( "dcb
\n" ); document.write( "-----\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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