2024/048) Show that if p is prime then ${2p}\choose{p}$$\equiv 2 \pmod p$

 We have

 ${2p}\choose{p}$$=\frac{(2p)!}{p!p!}$ bu definition

$=\frac{p!\prod_{k=p+1}^{2p}k}{p!p!}$ bu expansion

$=\frac{\prod_{k=p+1}^{2p}k}{p!}$ cancelling p! from both numerator and denominator

$=\frac{2p\prod_{k=p+1}^{2p-1}k}{p!}$ by taking 2 p ouut

working in mod p we get $(p+n) \equiv n \pmod p$

So we get

$=\frac{2p\prod_{k=p+1}^{2p-1}k}{p!} \pmod p$

$=\frac{2p\prod_{k=1}^{p-1}k}{p!} \pmod p$

$=2\frac{p\prod_{k=1}^{p-1}k}{p!} \pmod p$

$=2\frac{p!}{p!} \pmod p$

$=2$ 

Proved