A carton of 12 rechargable batteries contains one that is defective. In how many ways can the inspector choose 3 of the batteries and (a) get the one that is defective, (b)not get the one that is defective?

Question

anonymous · Answer

Can't seem to figure this one out. They get 55 for the first one.

anonymous · Answer

i guess for the first one .. that now we left with 11 batteries of which we have to take 2 of them .. 
so $$\left(\begin{matrix}11 \ 2\end{matrix}\right)$$

anonymous · Answer

oh okay, so $$\left(\begin{matrix}11 \ 2\end{matrix}\right)=\frac{10*10}{2*1}$$

anonymous · Answer

no $$\left(\begin{matrix}11 \ 2\end{matrix}\right) = 11! / (2! * 9!)$$

anonymous · Answer

Oh so its $$\left(\begin{matrix}n \ r\end{matrix}\right)=\frac{n!}{r!(n-r)!}$$

anonymous · Answer

yes

anonymous · Answer

Okay so I think I've figured out how they got the second one:
$$\left(\begin{matrix}11 \ 3\end{matrix}\right)=\frac{11!}{3!}=\frac{11*10*9}{3*2*1}=165$$

anonymous · Answer

So the restriction in this example,r, is 3 because I have three possibilities (by choosing three batteries at a time) of not getting the defective battery, where as in the previous scenario I had two possibilities . Is this correct?

anonymous · Answer

yes it is :)