# Pizza permutations problem

1. Jun 20, 2005

### himurakenshin

pizza store has small/medium/large with 10 different toppings 2 crusts and 3 types of sauses. how many ways to ordera pizza with atleast 1 topping and 1 sauce?

2. Jun 20, 2005

### whozum

This is straight nPk stuff.

3. Jun 22, 2005

### HallsofIvy

Staff Emeritus
This isn't really a "permutations" problem since order is not important. Use the "fundamental counting principal". If you have n choices for "X" and m choices for "Y", then you have mn choices for both X and Y together.

4. Jun 22, 2005

### LittleWolf

Does the "at least 1 topping and 1 sauce" mean AT LEAST ONE TOPPING AND ONLY ONE SAUCE or does it mean AT LEAST ONE TOPPING AND AT LEAST ONE SAUCE?

5. Jun 22, 2005

### geosonel

it means typical pizza of 1 sauce plus at least 1 topping.
for each of 3 sizes, each of 2 crusts, and each of 3 sauces, you'll need to consider:
(number ways choosing 1 topping
+ number ways choosing 2 toppings
+ number ways choosing 3 toppings
+ ... ... ... number ways choosing 10 toppings)

6. Jun 22, 2005

### honestrosewater

LittleWolf is right; The structure of the statement is ambiguous, and the meaning doesn't help much. You can certainly have more than one type of sauce on a pizza, just as you can have more than one topping. Just choose one meaning and solve for it, or solve for both meanings.

7. Jun 23, 2005

### gunblaze

This is a combination qn..Juz take 3*10*2*3

8. Jun 24, 2005

### Dr.Brain

You havent taken the condition into consideration.

9. Jun 26, 2005

### geosonel

# orders for each of 3 sizes, each of 2 crusts, at least 1 of 10 toppings, and at least 1 of 3 sauces =
(3 sizes)(2 crusts)

(number ways choosing 1 topping from 10
+ number ways choosing 2 toppings from 10
+ number ways choosing 3 toppings from 10
+ ... ... ... number ways choosing 10 toppings from 10)

(number ways choosing 1 sauce from 3
+ number ways choosing 2 sauces from 3
+ number ways choosing 3 sauces from 3)

$$= \ 6 \cdot \left ( \sum_{r=1}^{10} \mathbb{C}_{r}^{10} \right ) \cdot \left ( \sum_{r=1}^{3} \mathbb{C}_{r}^{3} \right )$$

$$= \ 6 \cdot \left ( \, (-1) \ + \ \sum_{r=0}^{10} \mathbb{C}_{r}^{10} \right ) \cdot \left ( \, (-1) \ + \ \sum_{r=0}^{3} \mathbb{C}_{r}^{3} \right )$$

$$= \ 6 \cdot \left ( \ (-1) \, + \, (2^{10}) \ \right ) \cdot \left ( \ (-1) \, + \, (2^{3}) \ \right )$$

$$= \ 6 \cdot \left (1023 \right ) \cdot \left (7 \right )$$

$$= \ 42,966$$