Question 1083305
The P's can have 
C( 5,2 ) = {{{ 5! / ( 2! * ( 5-2 )! ) }}}
C( 5,2 ) = {{{ 5! / ( 2! * 3! ) }}}
C( 5,2 ) = {{{ ( 5*4 ) / 2 }}}
C( 5,2 ) = {{{ 10 }}} combinations
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For each of these combinations, there are
P( 3,3 ) = {{{ 3*2*1 }}}
P( 3,3 ) = {{{ 6 }}} permutations
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C( 5,2 ) * P( 3,3 ) = {{{ 10*6 }}}
{{{ 10*6 = 60 }}}
There are 60 different ways
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Definitely get a 2nd opinion on this