Question

A carton of 12 rechargable batteries contains two that are defective. In how many ways can the inspector choose 3 of the bateries and get (a) none of the defective batteries, (b) one of the defective batteries, (c) both of the defective batteries?

Answer 1

Starting with a), I ask you, do you see that you are trying to see how many ways you can pick 3 things out of 10?

Answer 2

yes

Answer 3

sorry I meant to post: \[\left(\begin{matrix}10 \\ 3\end{matrix}\right)=\frac{10!}{3!}=120\]

Answer 4

You are right about the answer being 120, but it is represented by 10! / (7!3!), not 10! / 3!

Answer 5

To see this and to go forward with the rest of the problem, 10! / (7!3!) = (10 x 9 x 8) / (3 x 2 x 1). Do you see the difference?

Answer 6

Yes but what happens to the 7!?

Answer 7

The factor of 7! in the denominator serves to limit the numerator to 3 numbers (10 x 9 x 8) since you are picking only 3, hence 3 numbers. If this were a questions of permutations (which it is not, it is a question combinations), we would stop with 10x9x8 = 720 as our answer, but we don't care about the order, so we have to divide by 3x2x1 so as to eliminate the "multiple counting" of the ways.

Answer 8

oh okay.

Answer 9

In other words, there are 10 good batteries. We make a first pick, which could be any of the 10, call it G7. Now we have 9 left, so so far we have G7 and say G9. 8 left. We pick G6 for our third. 10x9x8, but G7G9G6 is the same as G9G6G7 and so forth, so we have to divide by 3x2. Still with me?

Answer 10

Yes

Answer 11

Ok, good, so on to part b. One of the batteries picked will be defective, which means that the other 2 will be good, so you have to see how many ways you can pick 2 good from 10 while at the same time 1 bad from 3. This is was it going on conceptually. Now you make a stab at what you think would be the way to set up the problem algebraically and check your thinking on this one.

Answer 12

That is I'll check your thinking on this one. Some words I'm typing here are not coming out.

Answer 13

This is what I get\[\left(\begin{matrix}10 \\ 2\end{matrix}\right) = \frac {10!}{8!2!}=45\]

Answer 14

Choosing one good battery from 3 gives me two chances to get a bad battery, so my r would be 2.

Answer 15

That is close, but not it. You are correctly thinking and accounting for the good batteries, but you have left out of your consideration the bad battery. You have to take into account the ways of picking 2 good ones while at the same time, any one of the 3 bad ones.

Answer 16

Think of the 2 good batteries as a bundle and now you have to get a bad one, so take 45 and use it along with the bad battery to be picked. I can't tell you what to add or multiply because I can't just give a number.

Answer 17

^^ In your first post above I think you meant to say "any one of the two bad ones"?

Answer 18

Yes, you're right. There are only 2 bad ones, so it's 45 along with something you have to do with the 2 possible bad ones to get some number.

Answer 19

I'm thinking that I would have to use the multiplication rule.

Answer 20

Yes, that's right. You can have that "bundle" of 2 good batteries (keep in mind the number that goes with it) with any of the 2 bad ones. So, you're almost there.

Answer 21

So by having choosen another good battery (2 out of the three set) I am left with the possibility of choosing one bad one so r would be 1 and the combination \[\left(\begin{matrix}10 \\ 1\end{matrix}\right)\]?

Answer 22

You have 45 for the "bundle" and 2 ways to get the "bad", so it's 45 x 2

Answer 23

oh ok.

Answer 24

So, part a was 120, part b was 90, and you just have to come up with part c.

Answer 25

So would this one be \[\left(\begin{matrix}10 \\ 1\end{matrix}\right)=\frac{10!}{9!1!}=10\]

Answer 26

The formula is not displaying on my screen for some reason. You might just have to ask about the resultant integer.

Answer 27

I'm thinking that I would have to follow the same steps that I did in the previous problem. So now I'm trying to find out how many ways I can get two bad from three and two good from ten.

Answer 28

So I would need to find two selections again.For r in my first combination, I got 1, and since ten is my n I got 10! over 9!1!=10

Answer 29

Actually, you are looking to see how many ways you can get 2 bad from 2 and 1 good from 10.

Answer 30

You are right. part c is 10 and you are finished!

Answer 31

Great, I think I'm starting to understand this a little bit better now. Thank you tcarroll

Answer 32

you're welcome.