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Math Help - Number of possible pizas.

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    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?
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    Quote Originally Posted by calculuskid1 View Post
    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.
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    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.
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    Quote Originally Posted by pickslides View Post
    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..
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    Quote Originally Posted by undefined View Post
    That assumes the order in which the toppings are placed is significant..
    Yep, but I'm not sure why that would ever matter!
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    Quote Originally Posted by pickslides View Post
    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)
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    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.}
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