# Thread: Number of possible pizas.

1. ## Number of possible pizas.

How many different varieties pizza can be made if you have the following choices: small, medium, or large size; thin n' crispy, hand-tossed, or pan crust, and 12 toppings (cheese is automatic), from which you may select from 0 to 12?
So I know theres 3 sizes of pizza, 3 types and 13 toppings but I dont know what to do with all of it?

2. Originally Posted by calculuskid1
How many different varieties pizza can be made if you have the following choices: small, medium, or large size; thin n' crispy, hand-tossed, or pan crust, and 12 toppings (cheese is automatic), from which you may select from 0 to 12?
So I know theres 3 sizes of pizza, 3 types and 13 toppings but I dont know what to do with all of it?
Not all that tricky when you think about it, also it's not quite accurate to interpret 0 to 12 toppings as 13 toppings.. try to see why it's 3*3*2^12.

3. hmmm, if cheese is a given and not included in the 12 then the number of combinations is as follows.

0 toppings: $3 \times 3 = 9$

1 topping: $3 \times 3 \times 12= \dots$

2 toppings: $3 \times 3 \times 12 \times 11 = \dots$

3 toppings: $3 \times 3 \times 12\times 11\times 10= \dots$

and so on.... (hoping you can see the pattern!)

Then sum all of these.

4. Originally Posted by pickslides
hmmm, if cheese is a given and not included in the 12 then the number of combinations is as follows.

0 toppings: $3 \times 3 = 9$

1 topping: $3 \times 3 \times 12= \dots$

2 toppings: $3 \times 3 \times 12 \times 11 = \dots$

3 toppings: $3 \times 3 \times 12\times 11\times 10= \dots$

and so on.... (hoping you can see the pattern!)

Then sum all of these.
That assumes the order in which the toppings are placed is significant..

5. Originally Posted by undefined
That assumes the order in which the toppings are placed is significant..
Yep, but I'm not sure why that would ever matter!

6. Originally Posted by pickslides
Yep, but I'm not sure why that would ever matter!
Well for what it's worth, if the order in which toppings are placed matters, then we get as you wrote

$\displaystyle 3\cdot3\cdot\sum_{k=0}^{12}P(n,k)=3\cdot3\cdot\sum _{k=0}^{12}\binom{n}{k}\cdot k!$

whereas if order doesn't matter it's

$\displaystyle 3\cdot3\cdot\sum_{k=0}^{12}\binom{n}{k}=3\cdot3\cd ot2^{12}$

(but we can jump to 2^12 without thinking of sums because every topping is either on or off, giving 2*2*..*2 = 2^12)

7. Hello, calculuskid1!

How many different varieties pizza can be made if you have the following choices . . .

Size: small, medium, or large
Style: "Thin n' Crispy", "Hand-tossed", or "Pan Crust"
Toppings: 12 choices

There are: . $\begin{Bmatrix} 3\text{ choices of size} \\ 3\text{ choices of style} \\ 2^{12}\text{ choices of toppings} \end{Bmatrix}$

There are: . $3 \times 3 \times 2^{12} \;=\;36,\!864\text{ varieties.}$