I have two pizzas, I can choose up to 3 topping, there are 7 toppings to choose from.
Double toppings, triple toppings of the same kind are allowed.
Order doesn't matter, ham on top of chicken is same as chicken on top of ham, so count as 1 pizza.
Question is how many ways can we order the pizza.
Here is what i have so far:
1 pizza 0 topping: $1$ way
1 pizza 1 topping: $7$ ways
1 pizza 2 toppings: $\binom{7}{2}$ unique topping + $7$ ways of picking same topping twice
1 pizza 3 toppings: $\binom{7}{3}$ unique topping + $7$ ways picking three of the same topping + $42$ $(7*2)$ways of picking two of the same + one unique
2 pizza 3 topping = $(1+7+\binom{7}{2}+\binom{7}{3}+7+7+42)^2 = 14400$
Is this correct? If so is there a quicker way and easier way of solving this problem?
Best Answer
You are correct in how many ways there are to make $1$ pizza. There are multiple ways to solve, but $120$ is the correct answer (your calculation is correct).
However, your mistake is when you go to two pizzas. If there are $120$ ways to order a pizza, then there are $$ \binom{120}{2}+120=\binom{121}{2}=7260 $$ ways to order two pizzas.
Consider ordering one pizza with chicken ($1$ topping) and one pizza with pepperoni ($1$ topping). This is no different than ordering one pizza with pepperoni ($1$ topping) and one pizza with chicken ($1$ topping). My counting counts this combination once, but yours counts them seperately.