counting question

[yamazaki]

I've been trying to figure out how to solve this problem, but with no luck

the question is :

you have 12 batteries 2 of them are defective

part A :In how many way can you choose 3 out of the 12 batteries and get 1 defective battery ?
part B :In how many way can you choose 3 out of the 12 batteries and get the 2 defective batteries ?

what i have done so far is this :
since we have 12C3 way to choose 3 out of 12
and there are 10C3 way to not get the defective ones

we have 12C3 - 10C3 = 100 ways to get the 2 defective batteries (that is for part B, i couldn't solve part A )

but according to the solution manual the right answers are:
A : 2 x 10C2
B : 10C1

please can someone explain this to me .. my head is gonna explode.
thank you in advance!

[Dason]

Your solution for A isn't correct because it's asking specifically for grabbing only one defective. Your solution allows for grabbing both defectives as well. To answer this you could reason that you need to grab only one defective so there are 2c1 ways to do that. Then you need to grab two non defectives so there are 10c2 ways to do that. Multiply them and you get the solution.

For part B you know you need both batteries and there are 2c2 = 1 ways to get both and then you need to select one of the non defectives and there are 10c1=10 ways to do that. Multiply together and you have your solution.

[yamazaki]

Thank you !!!

i was hitting my head against the wall and pull off my hair while trying to make sense out of this simple problem, WHY in the name of god there is no such explaining in my textbook ?!