Combination Problem (Please help)

Janlloyd

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I've been stuck on this specific problem, "John's Pizzeria offers 3 choices of crust, 3 choices of cheese and 10 choices of toppings. If topping choice can be repeated, how many distinct pizzas with crust, one cheese and two toppings can be ordered?" Ive gotten as far as to do 3C1 * 3C1 *, but after that I go blank because of the specific wording of "If topping choice can be repeated" so I don't know if to use 10C2 20C2 or 10C2 * 2. The answer choices are 702, 405, 330, 165, 495. Thank you very much for your help.
 
I've been stuck on this specific problem, "John's Pizzeria offers 3 choices of crust, 3 choices of cheese and 10 choices of toppings. If topping choice can be repeated, how many distinct pizzas with crust, one cheese and two toppings can be ordered?" Ive gotten as far as to do 3C1 * 3C1 *, but after that I go blank because of the specific wording of "If topping choice can be repeated" so I don't know if to use 10C2 20C2 or 10C2 * 2. The answer choices are 702, 405, 330, 165, 495. Thank you very much for your help.

There's a little ambiguity in the question, but if we take it to mean that you can either choose two different toppings, or two of the same, then you can count the number of ways to do each of these, and add them. That would give the number of ways to choose the toppings; then multiply by your other two numbers to get the total.
 
There's a little ambiguity in the question, but if we take it to mean that you can either choose two different toppings, or two of the same, then you can count the number of ways to do each of these, and add them. That would give the number of ways to choose the toppings; then multiply by your other two numbers to get the total.

So...

(3 kinds of crust to choose from) x (3 kinds of cheese to choose from ) x [ (the ways to choose 2 different toppings from 10 = 10C2 = 45) + (the ways to choose 2 of the same toppings from 10 = 2 x10) ] = 585?

Either that doesn't work or I did it wrong because that isn't one of Janlloyd's answer choices... hmmm...

Guiltily, I resort to test taking strategies... I'm fairly confident that starting out with 3C1 * 3C1 is correct. The equals 9. So, the final answer choice must be divisible by 9. That allows me to narrow down the answer choices to 702, 405, and 495. If I divide each of those possible answers by 9... I get 78, 45, and 55.

So what does it mean that "topping choice can be repeated?"

Would it mean you have to take two toppings and you can take any two toppings? So would that be 10^2? Well, there's no answer close to 900, so that can't be right. What if we divide out the times we double count instances of topping choice. for instance, choosing topping 1 and topping 2 is the same as choosing topping 2 and topping 1. By my count the would eliminate (1+2+3+4+5+6+7+8+9) = 45 choices. That leaves 55 choices. (100-45= 55. The way that this addresses the wording of the question is by counting all of the times that you get double pepperoni or double onion or whatever topping your decide to "repeat." However, it avoids counting pepperoni onion and onion pepperoni as separate outcomes because they aren't.

9 Times 55 equals 495. So, I am guessing that the answer is 495.

Janlloyd, is that correct? I REALLY WANT TO KNOW!!!!!!! LOL ;-)

Dr. Peterson, is there an easier way to do this?
 
So...

(3 kinds of crust to choose from) x (3 kinds of cheese to choose from ) x [ (the ways to choose 2 different toppings from 10 = 10C2 = 45) + (the ways to choose 2 of the same toppings from 10 = 2 x10) ] = 585?

Either that doesn't work or I did it wrong because that isn't one of Janlloyd's answer choices... hmmm...

Guiltily, I resort to test taking strategies... I'm fairly confident that starting out with 3C1 * 3C1 is correct. The equals 9. So, the final answer choice must be divisible by 9. That allows me to narrow down the answer choices to 702, 405, and 495. If I divide each of those possible answers by 9... I get 78, 45, and 55.

So what does it mean that "topping choice can be repeated?"

Would it mean you have to take two toppings and you can take any two toppings? So would that be 10^2? Well, there's no answer close to 900, so that can't be right. What if we divide out the times we double count instances of topping choice. for instance, choosing topping 1 and topping 2 is the same as choosing topping 2 and topping 1. By my count the would eliminate (1+2+3+4+5+6+7+8+9) = 45 choices. That leaves 55 choices. (100-45= 55. The way that this addresses the wording of the question is by counting all of the times that you get double pepperoni or double onion or whatever topping your decide to "repeat." However, it avoids counting pepperoni onion and onion pepperoni as separate outcomes because they aren't.

9 Times 55 equals 495. So, I am guessing that the answer is 495.

Janlloyd, is that correct? I REALLY WANT TO KNOW!!!!!!! LOL ;-)

Dr. Peterson, is there an easier way to do this?

Why do you think there are 20 ways to choose 2 of the same topping? There are only 10 such choices possible, since order doesn't matter. So you get 9(45 + 10) = 495, which is in the list.
 
my mistake

Why do you think there are 20 ways to choose 2 of the same topping? There are only 10 such choices possible, since order doesn't matter. So you get 9(45 + 10) = 495, which is in the list.


Ahh... my mistake. sorry. 9(45+10) would be the correct way to work that out.

I think writing it out like this would also lead to the correct answer...but I am not sure.

1,1 XX XX XX
1,2 2,2 XX XX
1,3 2,3 3,3 XX
1,4 2,4 4,4
1,5 2,5
1,6 2,6
1,7 2,7
1,8 2,8
1,9 2,9
1,10 2,10

and so on up to 10,10, where the numbers represent the 10 different toppings and the ordered pairs represent the possible combinations.
 
Ahh... my mistake. sorry. 9(45+10) would be the correct way to work that out.

I think writing it out like this would also lead to the correct answer...but I am not sure.

1,1 XX XX XX
1,2 2,2 XX XX
1,3 2,3 3,3 XX
1,4 2,4 4,4
1,5 2,5
1,6 2,6
1,7 2,7
1,8 2,8
1,9 2,9
1,10 2,10

and so on up to 10,10, where the numbers represent the 10 different toppings and the ordered pairs represent the possible combinations.

I can't follow what you are thinking there; but often the way to solve these is to imagine a way of literally counting possibilities, which then turns into a sequence of calculations that you can multiply or add; so some picture like yours may well lead to a good method. There are many ways to solve this one, including the subtractive method you used, where you eliminated 45 duplicates.

In fact, combinatorics becomes fun when you try to find alternate methods for any given problem; that's the way to strengthen your skills, never being satisfied having just one solution. (In complicated problems, I don't trust an answer until I can get it two ways.)
 
Thank you all

Thank you everyone for adding your input. It now makes sense to me that I had to add 10 to 10C2 to account for the tricky wording of "toppings can be repeated". The answer was indeed 495!
 
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