Question 1149892
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"51 choose 3"<br>
{{{C(51,3) = (51*50*49)/(3*2*1) = 20825}}}<br>
Anyone of the 51 can be chosen first; any of the remaining 50 second; and any of the remaining 49 third.  So 51*50*49 different ORDERS in which 3 of the 51 can be chosen.<br>
But for any given group of 3, the number of different orders in which they could have been chosen is 3*2*1.<br>
So the number of different COMBINATIONS of 3 chosen from 51 is (51*50*49)/(3*2*1).<br>