Hi,

If p is a prime number and r = 1,2,3 ...p -1, how can I prove that the binominal coeffisients nCr(p,r) are divisible by p? By divisible I mean nCr(p,r)/p leaves remainder 0.

oivind

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- May 25th 2012, 05:01 AMoeivinrdivisibility of binominal coeffisients
Hi,

If p is a prime number and r = 1,2,3 ...p -1, how can I prove that the binominal coeffisients nCr(p,r) are divisible by p? By divisible I mean nCr(p,r)/p leaves remainder 0.

oivind - May 25th 2012, 06:02 AMPlatoRe: divisibility of binominal coeffisients
- May 25th 2012, 08:26 AMoeivinrRe: divisibility of binominal coeffisients
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. - May 25th 2012, 08:45 AMPlatoRe: divisibility of binominal coeffisients
Look at what you posted.

nCr(p,r) [/SIZE]are divisible by p. Where does n come from? You said nothing about it anywhere. Now you need to learn to post what you mean.

If then . That is what you put in your OP.

If that is not the question then why did you post it that way? - May 25th 2012, 09:41 AMoeivinrRe: divisibility of binominal coeffisients
Sorry, my notation was a bit sloppy. What I meant by nCr was "n choose r" from combinatorics. Perhaps I should have used the alternative notation , C(p,r), instead. The number of combinations of selecting several things, r, out of a larger group. p.

- May 25th 2012, 10:15 AMPlatoRe: divisibility of binominal coeffisients