How to find the remainder of $\binom{2013}{101}$ when it is divided by 101

Question

I start first from the definition of $$\binom{n}r=\frac{n!}{r!(n-r)!} $$ then I used the Wilson's theorem for p is prime $$(p-1)!\equiv-1\pmod p$$ now how we can continue??

Answer 1

We have that $2013=19\cdot 101+94$, hence by Lucas' theorem

$$\binom{2013}{101}\equiv \binom{19}{1}\cdot\binom{94}{0}\equiv\color{red}{19}\pmod{101}.$$

Answer 2

It is better if you use this formula instead: $$\binom{2013}{101}=\frac{2013\cdot2012\cdot\ldots 1913}{101!}$$

The denominator is obviously a multiple of $101$, but not of $101^2$. Since $101^2>2013$, the same can be said about the numerator.

Simplify the factor $101$ from both, and you get $100!$ below. Above is a bit more complicated.

The only multiple of $101$ is $1919$, which is $19\cdot 101$. The remaining numbers form a complete residue system modulo $101$, so their product is congruent to $100!$, which cancels with the denominator.

Thus, the answer is $19$.