Question

show that \(\large \binom{2n}{n}\) is divisible by \(2\) for all \(n\ge 1\)

Answer 1

\[\left(\begin{matrix}2n \\ n\end{matrix}\right) = \frac{ (2n)! }{ n!n! }\]

Answer 2

nvm

Answer 3

Hmmm, I guess:\[
{2n\choose n} = {2n-1\choose n-1} + {2n-1 \choose n} = 2 {2n-1 \choose n}
\]

Answer 4

wow! pascal's rule does the job is it

Answer 5

Lol that's actually pretty clever

Answer 6

I got excited about this problem because it has many solutions
but what wio has is the most clever+short+simple one i think

Answer 7

So is that also skew symmetry

Answer 8

No, it isn't

Answer 9

whats skew symmetry ?
i do see the left and right ends are symmetrical

Answer 10

You could say: \[
f(k,m) ={k+m\choose k} = {m+k \choose m} = f(m,k)
\]A function defined this way would be symmetric, kinda

Answer 11

It's not skew symmetry, but I do see symmetry obviously, skew symmetry is for a matrix (I actually learnt that yesterday from empty) haha...thought I could apply it at other places as well

Answer 12

\[aij = −aji\] xD

Answer 13

Skew symmetry is more like subtraction, I suppose: \[
f(x,y) = x-y = -(y-x) = -f(y,x)
\]

Answer 14

Yeah you're right

Answer 15

But it's better to say 'commutative' than 'symmetric' when talking about maps, and to say 'symmetric' when talking about relations.

Answer 16

\[
f(k,m) ={k+m\choose k} = {m+k \choose m} = f(m,k)
\]

Again this follows from identity right
\[\binom{n}{r} = \binom{n}{n-r}\]

Answer 17

thats an interesting way to see it