Prove that $\gcd({n \choose i},{n \choose j})>1,~1

Question

Prove that $\gcd({n \choose i},{n \choose j})>1,~0<i,j<n$

My work:
I tried expanding $n \choose i$ and $n \choose j$ to find that there is some number that divides both and after division the numbers are still integers, but I could not prove that they are integers. Please help.

$(i,j) \neq 0,n$. I get what you mean but this is not correct notation. — user133281, Mar 07 '14 at 11:44
Please feel free to edit the question to introduce correct notation. — Hawk, Mar 07 '14 at 11:45
As stated this is not true. Set $i=j=n$ to get a counterexample — Amr, Mar 07 '14 at 11:46
@user133281 I think the edit you suggested was incorrect. I think you should re-edit it correctly to represent it in the desired form. — Hawk, Mar 07 '14 at 11:50
@Amr, Yes, I agree with your suggested edit, I will represent it like that.' — Hawk, Mar 07 '14 at 11:51

score 4 · Answer 1 · answered Mar 07 '14 at 13:13

Hint: Suppose that their greatest common divisor is $1$. Then the fraction $$\frac{{n\choose j}}{{n\choose i}}$$ is written in lowest terms.

On the other hand, this fraction can also be written as $$\frac{{n-i\choose n-j}}{{j\choose i}}\ldots$$

Prove that $\gcd({n \choose i},{n \choose j})>1,~1

1 Answers1