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](https://tex.z-dn.net/?f=C%2818%2C2%29%3D%20%5Cdfrac%7B18%21%7D%7B2%21%2818-2%29%21%7D%20%3D%5Cdfrac%7B18%21%7D%7B2%21%2816%29%21%7D%3D%5Cdfrac%7B17%5Ccdot%2018%7D%7B2%7D%3D153)
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](https://tex.z-dn.net/?f=C%2818%2C3%29%3D%20%5Cdfrac%7B18%21%7D%7B3%21%2818-3%29%21%7D%20%3D%5Cdfrac%7B18%21%7D%7B3%21%2815%29%21%7D%3D%5Cdfrac%7B16%5Ccdot%2017%5Ccdot%2018%7D%7B2%5Ccdot%203%7D%3D816)
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](https://tex.z-dn.net/?f=C%2818%2C4%29%3D%20%5Cdfrac%7B18%21%7D%7B4%21%2818-4%29%21%7D%20%3D%5Cdfrac%7B18%21%7D%7B4%21%2814%29%21%7D%3D%5Cdfrac%7B15%5Ccdot%2016%5Ccdot%2017%5Ccdot%2018%7D%7B2%5Ccdot%203%5Ccdot%204%7D%3D3060)
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 <span>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 </span>
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![C(18,0)+C(18,1)+C(18,2)+\dots+C(18,17)+C(18,18)=2^{18}](https://tex.z-dn.net/?f=C%2818%2C0%29%2BC%2818%2C1%29%2BC%2818%2C2%29%2B%5Cdots%2BC%2818%2C17%29%2BC%2818%2C18%29%3D2%5E%7B18%7D)
different ways of ordering.
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