#1

Show nCk = (n-1)Ck + (n-1)C(k-1)

that's read n choose k

it has something to do with the defitiion of nCk = n!/(k!(n-k)!)

any ideas?

that's read n choose k

it has something to do with the defitiion of nCk = n!/(k!(n-k)!)

any ideas?

#2

It's probably not the best idea to ask for math help in the pit.

#3

It's 11.

I might have to run through my work again to see if I made any mistakes.

Edit: Yeah, I made a few mistakes.

I might have to run through my work again to see if I made any mistakes.

Edit: Yeah, I made a few mistakes.

*Last edited by C O B H C at Feb 4, 2008,*

#4

IT R

#5

lawlz I know there are some brains out there

#6

I've tried a bunch, but I'll try again.

#7

Hows that not obvious??

nCk = (n-1)Ck + (n-1)C(k-1)

through this, we can conclude, that the inverse of (n-1)C(k-1) is C-2(nk) squared. Thus, we add C-2(nk)2 + (n-1)Ck to conclude that nk(ck)= -2(n-1)+(n+c). From here, we distribute the latter expression: (n+1)+(n+k-2)=nk(ck). So, n+k2+C=(n-1)C(k-1).

Thus, C=2, N=4, and K=6

now just substitute the numbers and, VIOLA!

nCk = (n-1)Ck + (n-1)C(k-1)

through this, we can conclude, that the inverse of (n-1)C(k-1) is C-2(nk) squared. Thus, we add C-2(nk)2 + (n-1)Ck to conclude that nk(ck)= -2(n-1)+(n+c). From here, we distribute the latter expression: (n+1)+(n+k-2)=nk(ck). So, n+k2+C=(n-1)C(k-1).

Thus, C=2, N=4, and K=6

now just substitute the numbers and, VIOLA!

#8

Sorry, I'm only in algebra 2

#9

#10

I flunked out of math

#11

it isn't algebra, it's a combination problem

#12

Um...

She's a woman?

She's a woman?

#13

^ well, aren't combination and permutation algebra..?

#14

42!

#15

may i sugest some icy hot on the balls?

that ussually gets my math juices flowin?

that ussually gets my math juices flowin?

#16

it isn't algebra, it's a combination problem

if you know so much why dont you have the answer?

exactly

#17

the answer is 12

#18

Show nCk = (n-1)Ck + (n-1)C(k-1)

that's read n choose k

it has something to do with the defitiion of nCk = n!/(k!(n-k)!)

any ideas?

I'll like to qoute my probability teacher "Do it by brute force"

let's start with

(n-1)Ck+(n-1)C(k-1)=(n-1)!/(k!(n-k)!)+(n-1)!/((k-1)!(n-k+1)!)

...

just add the fractions