[Math] Logic Pizza Toppings Ordering Question

combinationsdiscrete mathematicspermutations

So here's the question:

The menu at a pizza place offers 14 possible toppings from 3 categories. Customers circle the toppings that they want on the order pad. (The order of circles or order of toppings does not matter.)

Veggie (7): Mushrooms, Peppers, Onions, Olives, Capers, Artichoke, Pineapple
Meat (5): Salami, Pepperoni, Ham, Salmon, Anchovies
Cheese (2): Mozzarella, Feta

  1. In how many ways can a customer's order pad choose exactly one topping from each category?
  2. In how many ways can an order choose exactly two toppings from each category, if they aren't allowed to circle a topping twice (i.e no repeats)?
  3. In how many ways can an order pad choose up to two toppings from each category, if they aren't allowed to circle a topping twice? (No toppings are allowed).
  4. In how many ways can an order pad choose exactly two toppings from each category, i they are allowed to circle a topping twice (i.e, repeats allowed)?
  5. In how many different possible orders are there for Veggie-topping-only pizzas with at least two toppings, if repeated toppings are not allowed?

And here are my answers:

  1. C(7, 1) + C(5, 1) + C(2, 1)
  2. C(7, 1) + C(6, 1) + C(5, 1) + C(4, 1) + C(2, 1) + C(1, 1)
  3. C(7, 2) + C(5, 2) + (2, 2)
  4. (7 + 2 – 1)!/2!(7 -1)! (5 + 2 – 1)!/2!(5 -1)! (2 + 2 – 1)!/2!( -1)!
  5. C(7, 2)

Did I work the problems out correctly? If not, what's wrong with my answers?

My justifications:
1. For number 1, the question is asking for one topping from each category. The order doesn't matter so it's a combination problem. The first and second digit represent the total number of choices and the number of options you are allowed to pick, respectively. Thanks to the kind users in the comment section, I know understand that I was wrong to put in addition symbols, and that there should in fact be multiplication symbols in number 1.
2. This problem is very similar to number one except you can't repeat toppings. This means that the number of options you have to chose from the second time around are one less.
3. This one was rather tricky, and I doubt my answer here but I simply applied combination logic (i.e total number of choices, how many you are allowed to choose).
4. Here repetition is allowed so I picked the proper combination formula for that: (n + r - 1)!/r!(n-1)!
5. This problem asked only for the veggie topping pizzas with at least two toppings, so the answer was C(7, 2).

Best Answer

Partial answer

As said in comments the answer to $1)$ is $\binom{7}{1}\cdot \binom{5}{1} \cdot \binom{2}{1}.$

The answer to $2)$ is $\binom{7}{2}\cdot \binom{5}{2} \cdot \binom{2}{2}$ (you choose two vegetables and you have $\binom{7}{2}$ possible ways to do it, and similar for meat and cheese. Remember that you have to multiply).

The answer to $5)$ is

$$\binom{7}{2}+\binom{7}{3}+\binom{7}{4}+\binom{7}{5}+\binom{7}{6}+\binom{7}{7}.$$ If you choose two vegetables you have $\binom{7}{2}$ ways to do it, If you choose three vegetables you have $\binom{7}{3}$ ways to do it, and so on. Use that

$$\binom{7}{2}+\binom{7}{3}+\binom{7}{4}+\binom{7}{5}+\binom{7}{6}+\binom{7}{7}=2^7-\binom{7}{0}-\binom{7}{1}$$ to get its value.

Could you get the solution to the other questions?

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