# Boards

## someone who's good at maths, show me how this works.

This isnt cheating, its learning. Im not sure how im going to write this out but...

Here we go:

Prove that

(K+2) - (k)

(3) (3) = K^2

In english it would read

"Prove that "K+2 choose 3" minus "K choose 3" equals "K squared""

Cheers!

## what are you calling

K choose 3 ?

My mathematics english is a bit rusty...

## wtf does k choose 3 mean

## k over/divided by 3?

## apparently it has to do with factorials

I.e "5 choose 1" means "how many different combinations can you make with 1 person in a group of 5"

The answer is obviosuly 5.

You work it out by doing 5!/1!.4!

So I need to know basically how to do K!, but I cant

## ok. errrm

there must be a law/theorem in there somewhere.

## back in the zone again

I'm lost in a K! hole :(

## so, if I understand you correctly

k + 2 choose 3 = ( k + 2 )!/ ( k + 2 - 3 ) !. 3 ! = ( k + 2 )( k + 1 ) k / 6 = ( k^3 + 3k² + 2k )/6

while

k choose 3 = k! / ( k - 3 )!. 3! = k ( k - 1 )( k - 2 ) / 6 = ( k^3 - 3k² + 2k )/6

The rest is obvious.

## haha

I could'ave done it all along. I forgot to do the bottom line

Thanks lyle!

## wait...

why does ( k + 2 - 3 ) !. 3 ! = 6 ?

## i don't get why there is even a -3 in there

## because to get 5 choose 2

you do the first number factorial divided by the second number factorial times the factorial of the difference between the first and second numbers.

## what?

yuck.

## It doesn't

(k + 2)! = (k + 2)(k + 1)(k)(k - 1)...

aka (k + 2)(k + 1)(k)((k - 1)!)

The (k - 1)! cancels with the (k - 1)! on the bottom, leaving just 3! on the bottom, which is 6.

## Ah ha

makes sense.

## im so glad im not at school anymore

.