JOMO's Pizza

Probability Level 5

At JOMO's Pizzeria, you can design your own pizza. You can choose from any of the 6 toppings: Pepperoni, Salami, Onions, Mushrooms, Pineapples, and Ham. You can also choose between a thick and thin crust. A plain pizza with no toppings costs $120 HKD. Each extra topping you add on costs an extra $20 HKD. If I only have $200 HKD, how many different types of pizza could I order?

Details and Assumptions

You are allowed to order a Pizza with repeating toppings, so you could order a pizza with 3 toppings of all pineapple.

This problem was taken from JOMO.

The answer is 420.

1 solution

Yan Yau Cheng
Feb 16, 2014

This is Actually a question in JOMO 1, if you think this question wasn't too hard but isn't too easy, then you should take part in JOMO!

The Pizza I order can have a maximum of 4 toppings.

If there are no toppings, I can order 2 different pizzas, one with thick crust and one with thin crust.

With 1 topping, There are $C^6_1=6$ different toppings you can choose, and you can choose between thick and thin crust. So there are 12 different pizzas you can order with 1 topping.

With 2 toppings, you can have 2 different toppings or two identical toppings, which has $C^6_2 = 15$ and $C^6_1 = 6$ different combinations respectively. You can choose between thick and thin crust. So there are $(15+6)\times 2 = 42$ different pizzas you can order with 2 toppings.

With 3 toppings, you can have 3 different toppings, two identical toppings and 1 different, and 3 identical toppings, which has $C^6_3 = 20$ , $C^6_1C^5_1 = 30$ , and $C^6_1=6$ different combinations respectively. You can choose between thick and thin crust. So there are $(20+30+6)\times 2 = 112$ different pizzas you can order with 3 toppings.

With 4 toppings, you can have 4 different toppings, 2 identical toppings and 2 different, 2 identical toppings and another 2 identical toppings, 3 identical toppings and 1 different, and 4 identical toppings, which has $C^6_4 = 15$ , $C^6_1\frac{C^5_1C^4_1}{2!} = 60$ , $\frac{C^6_1C^5_1}{2!} = 15$ , $C^6_1C^5_1=30$ and $C^6_1=6$ different combinations respectively. You can choose between thick and thin crust. So there are $(15+60+15+30+6)\times 2 = 252$ different pizzas you can order with 4 toppings.

In total there are $2+12+42+112+252=\boxed{420}$ Different Pizzas I can order with $200 HKD

You can do this without casework. Ignore the different crusts to begin with. Also, denote the number of toppings of pepperoni as $a_{1}$ etc. up to ham as $a_{6}$ . So in fact we want to find the number of solutions to $a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} \le 4$ , where all terms are non-negative integers. This is also the same as the number of solutions to $a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + k = 4$ where again all terms are non-negative integers. Therefore, using stars and bars, the number of possibilities is $\dbinom{7+4-1}{4} = \dbinom{10}{4}$ . But each pizza can also either be thin crust or thick crust, meaning that the total number of possibilities is $2 \dbinom{10}{4} = \fbox{420}$

Josh Rowley - 7 years, 3 months ago

Isn't this also called the pigeon-hole principle?

Harshal Sheth - 7 years, 3 months ago

No. It is called Stars and bars . Josh's approach of adding an additional term to deal with the inequality is extremely useful to simplify the calculations.

You can learn more about the Pigeonhole Principle .

Calvin Lin Staff - 7 years, 3 months ago

I have just read about Stars and Bars and I thought it was explained well. But, I got confused with how from $a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + k = 4$ , you got ${7 + 4 - 1 \choose 4}$ . I am thinking that $7$ refers to the number of terms ( $a_1$ to $k$ ). Am I correct? And, where do the numbers ${7 \boxed{+ 4} \boxed{- 1} \choose 4}$ come from?

Ralph Anthony Espos - 7 years, 3 months ago

So you know that a1, ... k can be 0 to whatever. But stars and bars only works when they're greater than or equal to 1. So you let a1=a'-1 etc. so then you have a1'+a2'+a3'+...k'=7+4=11. So now you want stars and bars on a1' etc. and then when you do that, subtract 1 from everything to get your original a's and k's. There are 11-1 choose 7-1 = 10 choose 6 == 10 choose 4 ways.

faraz masroor - 7 years, 2 months ago

@Faraz Masroor – Er... not really the explanation I expected.

Ralph Anthony Espos - 7 years, 2 months ago

I came up first with $3110$ because I assumed that a pizza with $20$ HKD worth of, say, pineapple is different from another pizza with $40$ HKD worth of pineapple (because it had more toppings). I really wish the problem was clear with that.

Oh, well. Back to Level 3.

Ralph Anthony Espos - 7 years, 3 months ago

Me too :)

Tan Li Xuan - 7 years, 3 months ago

Me also.. The problem is vague about this. Personally, I still consider a pizza with 2 pineapple toppings as being of a different type than a pizza with 1 pineapple topping..

Rahul Choudhary - 7 years, 3 months ago

Same here

Sadhana Senthilkumar - 7 years, 2 months ago

With 2 toppings should not we have 2 6^2=72 with 3 toppings 2 6^3=432 and so on

Jayanta Mandi - 7 years, 3 months ago

Please refer to the Stars and Bars algorithm Calvin was above talking about, to clarify your doubt.

Soumya Chakraborty - 7 years, 3 months ago

This was among the few questions, I couldn't get on JOMO.

Shabarish Ch - 7 years, 3 months ago

JOMO's Pizza

This problem was taken from JOMO.

The answer is 420.

1 solution

0 pending reports