combination question: given nC2 = 6, how to solve for n

[brentwoodbc]

when you have a question like nC2=6
solve for n

n(n-1)(n-2)/2!(n-2)! = 6
=
n(n-1)/2 = 6
n^2-n+12=0
factor
(n+3)(n-4)=0
n=-3,4 **********n cant be negative so n=4
I dont know why n cant equal a negative number? I think there was another unit in my math class where x couldnt be negative (I cant remember)

I made up that question so so I dont have trouble solving it, I just want to know why it cant be negative.

thanks.

 

[soroban]

Re: combination question

$\displaystyle \text{Hello, brentwoodbc!}$

$\displaystyle \text{When you have a question like: }\:_nC_2 \,=\, 6.\;\; \text{ Solve for }n.$

$\displaystyle \frac{n(n-1)(n-2)!}{2!(n-2)!} \:=\: 6 \quad\Rightarrow\quad \frac{n(n-1)}{2} \:=\:6 \quad\Rightarrow\quad n^2-n+12\:=\:0$

$\displaystyle \text{Factor: }\;(n+3)(n-4)\:=\:0$

\(\displaystyle n\:=\:-3,\:4\quad \hdots\quad n\text{ can't be negative, so }n=4\)

$\displaystyle \text{I don't know why }n\text{ can't equal a negative number.}$

$\displaystyle \text{If }n = \text{-}3\text{ is a solution to the problem,}$
. . $\displaystyle \text{we can check our answer in the original equation . . . right?}$

$\displaystyle \text{So we would have: }\;_{\text{-}3}C_2 \:=\:6$

$\displaystyle \text{Then we'd have: }\:\frac{(\text{-}3)!}{^{something}} \:=\:6$

$\displaystyle \text{And a factorial of a }negative\text{ number is not defined.}$

 

[brentwoodbc]

Re: combination question

Ok thanks. Now I just have to remember what the other case was where x cant be a negative number....I think it was in the trig unit.