Combinatronics

chipjet wrote:A university cafeteria offers 4 flavors of pizza. If a customer has an option (but not the obligation) to add extra cheese, mushrooms, or both to any kind of pizza, how many different pizza varieties are available?

a) 4
b) 8
c) 12
d) 16
e) 32

Combinatronics? Is that one of them robots that turns into a calculator or something? I think you meant combinatorics

We have 4 base pizzas. For each of those pizzas, we could add:

1) nothing
2) extra cheese
3) shrooms
4) cheese and shrooms

So, each of the 4 pizzas has 4 options. 4 * 4 means 16 total possible pizzas: choose (d).

Ha! See, I don't even know what these problems are called! No wonder I have issues with them!

Okay, I see what you're saying. I thought about it like this. Why is this wrong?

Each pizza can be prepared in 4 ways: plain, extra cheese, mushrooms, both. So shouldn't we have:

4!+4!+4!+4!?

That would be the number of choices we would have... right? Where is the flaw in my logic?

Thanks!

chipjet wrote:Ha! See, I don't even know what these problems are called! No wonder I have issues with them!

Okay, I see what you're saying. I thought about it like this. Why is this wrong?

Each pizza can be prepared in 4 ways: plain, extra cheese, mushrooms, both. So shouldn't we have:

4!+4!+4!+4!?

That would be the number of choices we would have... right? Where is the flaw in my logic?

Thanks!

I agree that each pizza can be prepared in 4 different ways. I'm not sure why you're using factorials, though.

We have 4 pizzas, each of which can be prepared in 4 different ways. So the total number of possibilities is simply:

4 + 4 + 4 + 4 = 16