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Homework Help: Pizza permutations problem

  1. Jun 20, 2005 #1
    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. jcsd
  3. Jun 20, 2005 #2
    This is straight nPk stuff.
     
  4. Jun 22, 2005 #3

    HallsofIvy

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    Science Advisor

    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.
     
  5. Jun 22, 2005 #4
    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?
     
  6. Jun 22, 2005 #5
    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)
     
  7. Jun 22, 2005 #6

    honestrosewater

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    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.
     
  8. Jun 23, 2005 #7
    This is a combination qn..Juz take 3*10*2*3
     
  9. Jun 24, 2005 #8
    You havent taken the condition into consideration.
     
  10. Jun 26, 2005 #9
    based on previous thread contributions:
    # 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)

    [tex] = \ 6 \cdot \left ( \sum_{r=1}^{10} \mathbb{C}_{r}^{10} \right ) \cdot \left ( \sum_{r=1}^{3} \mathbb{C}_{r}^{3} \right ) [/tex]

    [tex] = \ 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 ) [/tex]

    [tex] = \ 6 \cdot \left ( \ (-1) \, + \, (2^{10}) \ \right ) \cdot \left ( \ (-1) \, + \, (2^{3}) \ \right ) [/tex]

    [tex] = \ 6 \cdot \left (1023 \right ) \cdot \left (7 \right ) [/tex]

    [tex] = \ 42,966 [/tex]
     
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