Hot dog combinatorics A hot dog stand has 12 different toppings available.  How many different kinds of hot dogs can be made, assuming the order of the toppings does not make a difference.  I believe the correct answer is 882050, with the maximum varieties per number of toppings selected being 665280 when there six toppings.  I am also not sure about how to create a formula that would arrive at this result.
 A: For each topping you can choose to include it or not include it.  This results in $2^{12}=4096$ different kinds of hot dogs.  
A: It's not necessarily a yes or no question on each topping.
For each topping, I would make 4 different 'states'


*

*none

*regular

*extra

*easy


$4^{12}=16,777,216$ individual hot dogs, which would definitely not fit in the stand. You could refine it more to have different standardized volumes of topping, but then you'd just get ridiculous. You could further confuse the issue by counting the bun as being a topping, because you wouldn't really have easy or extra bun, hun.
A: For each topping you can decide to use it or not to use it. 
So for each topping you have $2$ ways. Thus in total you have $2^{12}$ ways. 
A: Also, you can view it like this:
$$
\sum_{k=0}^{12} C_{12,k} = \sum_{k=0}^{12} \binom{12}{k} = 4096
$$
i.e. for each number k of toppings, you get k-combinations between those elements, from the starting 12. We are using combinations without repetition because order of selection does not matter.
Numerically:
$$\sum_{k=0}^{12} \binom{12}{k} = 1 + 12 + 66 +220 +495 +792 +924 +792 +495 +
220 +66 +12 +1 = 4096
$$
A: The vendor can make uncountably many different kinds of hot dogs but varying the amount of one single topping, if the amount can be expressed as a real number (e.g. weight).
