- #1

- 87

- 1

## Homework Statement

If p is a prime and k is an integer for which 0<k<p, then p divides [itex]\displaystyle \ \binom{p}{k}[/itex].

Whne p divides [itex]\displaystyle \ \binom{p}{k}[/itex] it means that [itex]\displaystyle \ \binom{p}{k}=p*b[/itex].

wheren b is some number.

## The Attempt at a Solution

So p is equal to some number k.

[itex]p=k[/itex]

I take factorial of both sides and get

[itex]p!=k![/itex]

and multiply both sides by

[itex]\frac{1}{k!(p-k)!}[/itex]

to get

[itex]\frac{p!}{k!(p-k)!}=\frac{k!}{k!(p-k)!}[/itex]

Since p=k

[itex]\displaystyle \ \binom{p}{k}=\frac{p(p-1)(p-2)...(p-k+1)(p-k)!}{k!(p-k)!}[/itex]

then

[itex]\displaystyle \ \binom{p}{k}=p*\frac{(p-1)(p-2)...(p-k+1)}{k!}[/itex]

where [itex]b=\frac{(p-1)(p-2)...(p-k+1)}{k!}[/itex] therefore p divides [itex]\displaystyle \ \binom{p}{k}[/itex].

--------------------------------------------------------------------------------------

## Homework Statement

If [itex]n \in N[/itex] then [itex]\displaystyle \ \binom{2n}{n}[/itex] is even i.e.

[itex]\displaystyle \ \binom{2n}{n}=2*b[/itex] where b is some number

## The Attempt at a Solution

[itex]\displaystyle \ \binom{2n}{n}=\frac{2n!}{n!(2n-n)!} = \frac{2n(2n-1)..3*2*1}{n!(2n-n)!}=2*\frac{n(2n-1)..3*2*1}{n!(2n-n)!}[/itex] where [itex]b=\frac{n(2n-1)..3*2*1}{n!(2n-n)!}[/itex]

Last edited: