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Suppose You are ordering two pizzas.A pizza can be small,medium,large or extra large,with any combination of 8 possible toppings( getting no toppings is allowed, as is getting all 8) How many possibilities are there for your two pizzas?

Answer:- My answer is $1,048,576$ possibilities

You can select any one of the four sizes in 4 ways. You can select any combination of 8 possible toppings in $(2)^8=256$ ways. For one pizza total possibilities are $4*256=1024$.So, for two pizzas, $(1024)^2=1,048,576$ possibilities.

If we treat sizes and toppings of two pizzas indistiguishable, then we can select 6 combinations of pizzas as per sizes.

These combinations are 1)Small-Medium 2)Small-Large 3)Small-Extra Large 4)Medium-Large 5)Medium-Extra Large 6)Large-Extra Large.

Each combinations can have 255*256=65280 different toppings from the 8 types of toppings. So the total possibilities for two pizzas are 3,91,680.

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    $\begingroup$ I want to know my answer is correct or not. I don't have the correct answer for confirmation. $\endgroup$ Commented Jun 9, 2019 at 15:47
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    $\begingroup$ First, try it for S and L pizzas with or without mushrooms & with or without sausage. $\endgroup$
    – BruceET
    Commented Jun 9, 2019 at 18:48
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    $\begingroup$ This is a [self-study] question, so please tell us how did you tried to solve it and where are you stuck? What kind of help you need from us? I see that you edited your question to delete the attempted solution, but it would really help if you've shown what have you tried, even if it is incorrect. $\endgroup$
    – Tim
    Commented Jun 10, 2019 at 6:35
  • $\begingroup$ @Tim, i was not confident about the previously attempted solution, so when i got correct solution after some logical thinking, i deleted the wrong solution and added correct solution. $\endgroup$ Commented Jun 10, 2019 at 7:14
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    $\begingroup$ Actually it depends if the pizzas are distinguishable or not. $\endgroup$
    – gunes
    Commented Jun 10, 2019 at 7:51

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So... If you only want to know whether the answer is right or wrong.... It's right! Good job!

But for the next time, consider using http://math.stackexchange.com since your question has nothing to do with statistics or probability

EDIT: As @gunes pointed out, you must divide by two if the order of the pizzas does not matter

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  • $\begingroup$ I think you need to try to explain why you'd divide by two - you'll probably see then why you wouldn't. $\endgroup$ Commented Jun 10, 2019 at 11:15
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    $\begingroup$ @David. If you roll two distinguishable dice, there are 36 possible rolls. However, if they are not distinguishable, and 1,2 is same as 2,1, then you have 21 possible rolls 1,1 2,2,..6,6 were each only counted once, so they weren't double counted. A good trick in these problems is to make the problem simple enough that you can enumerate the possibilities and check your logic. $\endgroup$
    – jsk
    Commented Jun 10, 2019 at 14:52
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    $\begingroup$ Or, sticking with pizza, if there are just two types, A & B, then the $2^2$ distinct ordered pairs are A-A, B-B, A-B, & B_A. How many distinct unordered pairs are there? $\endgroup$ Commented Jun 10, 2019 at 15:18
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    $\begingroup$ @David Do the math for the situation that scorchi described. By your logic, you would multiply 2×2=4 (squaring the possibilities for 1st and 2nd pizza) then divide by 2 to remove the duplicates leaves you with an answer of 2, which is wrong because the 3 distinct possibilities are AA, AB, and BB. You are correct that some possibilities are double counted, but wrong about how to remove the duplicates. $\endgroup$
    – jsk
    Commented Jun 10, 2019 at 16:33
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    $\begingroup$ I see - sorry! But is it not the case then that to get the number of distinct unordered pairs, you divide only the count of distinct ordered unalike pairs by two and add that to the count of distinct ordered like pairs? So with just two types of pizza, A & B, there are three distinct unordered pairs: A-A, B-B, & A-B? $\endgroup$ Commented Jun 10, 2019 at 16:37

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