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

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

any ideas?
GEAR

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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.
Last edited by C O B H C at Feb 4, 2008,
lawlz I know there are some brains out there
GEAR

VOX AC15 / AC4
Fender MIM Sunburst Strat
epiphone les paul custom in white
Yamaha FG730s
The Tweak-O fuzz pedal
I've tried a bunch, but I'll try again.
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!
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Sorry, I'm only in algebra 2
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it isn't algebra, it's a combination problem
GEAR

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Um...

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^ well, aren't combination and permutation algebra..?
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it isn't algebra, it's a combination problem

if you know so much why dont you have the answer?
exactly
Vikings? What Vikings? We are but poor, simple farmers. The village was burning when we got here, and the people must have slain themselves.
Quote by mju4t
Show nCk = (n-1)Ck + (n-1)C(k-1)

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"