Originally Posted by
oeivinr Thanx for answering. Of course p is a factor of p!, but not necessarily of nCr(p,r). To be a factor of nCr(p,r) it has to devide nCr(p,r) exactly, that is division by p will give an integer as an answer. e.g. nCr(11,3) =165 and 11 devides 165 exactly 165/11 = 15. This holds only for primes in the formulae nCr(p,r) not for composites e.g. nCr(6,2) = 15 but 6 does not divide 15 exactly 15/6 =2.5 I understand that r does not devide p. A prime is only divisible (exactly) by 1 and itself, but I don't quite see the relevance of this fact to my question.