The number of choices of 3 kinds of crust and up to 6 distinct toppings 
David has a pizza shop. There are 3 kinds of crust and 6 different toppings he can chose from. If customers can have as many toppings as they'd like but may not order double of one topping, how many different kinds of pizza can he make?

Simple enough but I can't figure it out. The answer is 192 but I don't know how to get to that. 

At first I thought it was a simple $3*6!$, but that yielded the wrong answer. I next figured since it said, "At least" then it would be a factorial with multiple stages added together like: $3\cdot 6+3\cdot 6\cdot 5+3\cdot 6\cdot4\cdots$ etc but once again it was wrong.
 A: Suppose you have n toppings and $m$ crusts then you can either say yes or no to a topping. So you have two choices for each topping and $m$ choices of crusts.
$$C =  m \times \underbrace{2}_{Yes \ or \ No} \times  \underbrace{2}_{Yes \ or \ No} \cdots \underbrace{2}_{Yes \ or \ No}\times  \underbrace{2}_{Yes \ or \ No} \times  \underbrace{2}_{Yes \ or \ No} = m \times 2^n$$
A: The number of ways you can order toppings is the number of subsets of a set with $6$ elements.  For each topping, you have two choices, include it or not include it.  Therefore, there are $2^6$ ways of ordering the toppings and three ways or ordering the crust, yielding $3 \cdot 2^6 = 192$ different types of pizza.
A: For a given crust there are seven pizzas possible (one with six toppings one with five toppings etc all the way to 0 toppings). To compute how many possible pizzas you could make with one crust would be consider how many combinations could be made from the 6 toppings if its a six topping pizza, how many can be made from the 6 toppings for a five topping pizza, etc. There are three crusts, so
$$\text{No. of Pizzas}=3\left(\binom{6}{6} + \binom{6}{5} + ... + \binom{6}{1}+\binom{6}{0}\right) = 3\times 64 = 192$$
Hope this helps 
