# Thread: divisibility of binominal coeffisients

1. ## divisibility 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

2. ## Re: divisibility of binominal coeffisients

Originally Posted by oeivinr
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.
If $\displaystyle p$ is a prime and $\displaystyle 1<r<p-1$ then because $\displaystyle r\not | ~p$. WHY?

Moreover, $\displaystyle p$ is always a factor of $\displaystyle p!$. So you are almost done.

Finish?

3. ## Re: 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.

4. ## Re: divisibility of binominal coeffisients

Originally Posted by oeivinr
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
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.

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.
If $\displaystyle r=2,~3,~\dots~10$ then $\displaystyle 11~|~\binom{11}{r}$. That is what you put in your OP.

If that is not the question then why did you post it that way?

5. ## Re: 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.

6. ## Re: divisibility of binominal coeffisients

Originally Posted by oeivinr
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.
$\displaystyle C(n,r)=\binom{n}{r}=\frac{n!}{r! (n-r)!}$ is the mathematical way to do that.

From that notation you can see that if $\displaystyle p$ is prime and $\displaystyle 1<r<p$ where $\displaystyle r\in\mathbb{Z}^+$
then it is true that $\displaystyle p$ divides $\displaystyle \binom{p}{r}$.

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# if n is a prime ncr is divisible by n

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