Question 1190906
this is 55C5=3,478,761 different committees
there are ultimately going to be 5 on the committee and 50 not.
There are 55*54*53*52*51*50! ways to choose the first
There are 5*4*3*2*1 ways to choose the 5 people on the committee
There are 50*49*48*47*46*....ways to choose the 50 people NOT on the committee
it is 55*54*53*52*51*50!/50!*120, the 120 being 5!, or 5*4*3*2*1