A pizza restaurant has 3 crust options, 2 cheese options and 10 choices of toppings. On Saturday nights, the restaurant offers a special deal on 2-toppings pizzas including pizzas with double portions of one toppings. How many distinct special deal pizzas are possible. My approach: I assumed (not too sure if this is correct) that a special deal has to contain a double portion. If this is the case, then $($crust$)\times($cheese$)\times($topping combinations$) =$ pizza combinations There are $\binom31$ of choosing a crust. $\binom21$ of choosing a cheese. There are $\binom<10>$ ways of choosing two toppings, but each for every pizza there is at least one double portion so there are $\binom<10> \times 2$ of choosing two toppings where a one of the toppings is a double size. This creates a total of $3\times2\times90=540$ ways of choosing a pizza. The answer is $330$ which could possible suggest that there are $\frac= 55$ ways of choosing a topping. How could this be? Where in my reasoning did I go wrong? Favorite pizza?
asked Nov 12, 2016 at 23:41 889 7 7 silver badges 25 25 bronze badgesThe key here is that you can select twice the same topping. (This is what "double portion" means here.)
So the number of topping combinations for a special deal is:
In total, you do get $\binom+10=55$ different possibilities for the choice of two toppings.