Question 565266
Pascal's triangle easily proves it, as


*[tex \LARGE nCr + nC(r+1) = (n+1)C(r+1)]


Since nC(r+1) is nonnegative, nCr <= (n+1)C(r+1).


Or, we can show that


*[tex \LARGE \frac{nCr}{(n+1)C(r+1)} \le 1]


*[tex \LARGE \frac{\frac{n!}{(n-r)!r!}}{\frac{(n+1)!}{(n-r)!(r+1)!}} \le 1]


Multiply both numerator and denominator by 1/(denominator)


*[tex \LARGE \frac{n!(r+1)!}{(n+1)!r!} \le 1 \Rightarrow n!(r+1)! \le (n+1)!r!]


Divide both sides by n!r!


*[tex \LARGE r+1 \le n+1 \Rightarrow r \le n]


This is true because we're choosing r objects out of n, so it is reasonable to assume r <= n.