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 $\binom{10}{2}$ ways of choosing two toppings, but each for every pizza there is at least one double portion so there are $\binom{10}{2} \times 2$ 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 $\frac{330}{6}= 55$ ways of choosing a topping.
How could this be? Where in my reasoning did I go wrong? Favorite pizza?
Best Answer
The 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 $\binom{10}{2}+10=55$ different possibilities for the choice of two toppings.