Question

The pizza restaurant down the street wants to advertise on its marquee sign the number of ways you can order a $10 pizza. The restaurant offers 18 toppings and for $10 you can order up to 4 toppings, meaning you can either order 1, 2, 3, or 4 toppings and you can’t order any toppings twice. Use this information to answer the following questions: a. Is this a permutation problem, a combination problem, or a problem using both elements? b. Determine how many ways you can order a pizza from the restaurant. Explain what strategies you used. c. If there were no restriction on the number of toppings you could choose, how would that change the answer?

Answer 1

1. You can order pizza with:  1, 2, 3, 4 toppings for $10. Let's count:

1 topping from 18: 18 ways;

2 toppings: $C(18,2)= \dfrac{18!}{2!(18-2)!} =\dfrac{18!}{2!(16)!}=\dfrac{17\cdot 18}{2}=153$ ways of ordering;

3 toppings:$C(18,3)= \dfrac{18!}{3!(18-3)!} =\dfrac{18!}{3!(15)!}=\dfrac{16\cdot 17\cdot 18}{2\cdot 3}=816$ ways of ordering;

4 toppings:$C(18,4)= \dfrac{18!}{4!(18-4)!} =\dfrac{18!}{4!(14)!}=\dfrac{15\cdot 16\cdot 17\cdot 18}{2\cdot 3\cdot 4}=3060$ ways of ordering.

Totally, 3060+816+153+18=4047 ways of ordering pizza with topping. Since you use here C(n,r) this is combinations problem.

 If there are no restriction on the number of toppings you could choose, you could order pizza with 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 toppings and totally it will be


$C(18,0)+C(18,1)+C(18,2)+\dots+C(18,17)+C(18,18)=2^{18}$ different ways of ordering.