Binomial Question... (2 Viewers)

FRIZZ_24 · Oct 24, 2014

How do you find n in:
nC3 - 2 n+1C2 = nC1

Thanks

integral95 · Oct 24, 2014

1C2 isn't defined

integral95 · Oct 24, 2014

I'll assume the question is nC3 - 2 n+nC2 = nC1

you need to use the formula $nCk = \binom{n}{k} = \frac{n!}{(k!)(n-k)!}$

so you would get $\frac{n!}{(3!)(n-3)!} + 2n + \frac{n!}{(2!)(n-2)!} = n$

SOlve for n there .

FRIZZ_24 · Oct 25, 2014

Thanks....but sorry for not being clearer - the q has (n+1)c2
I.e.

nC3 - 2 x (n+1)C2 = nC1

faisalabdul16 · Oct 25, 2014

Binomial Question... (2 Viewers)

FRIZZ_24

New Member

integral95

Well-Known Member

integral95

Well-Known Member

FRIZZ_24

New Member

faisalabdul16

Wot

Users Who Are Viewing This Thread (Users: 0, Guests: 2)