How many ways can three toppings be chosen from seven options?
(1) We assume that (i) you cannot repeat toppings and (ii) the order does not matter -- i.e. it doesn't matter whether you put sprinkles then nuts then chocolate syrup or nuts then syrup then sprinkles.
Under these assumptions there are three good methods to find the answer:
The counting principle -- you have 7 choices for the 1st topping, then 6 choices remaining for the second topping, then 5 choices for the last topping. Thus there are `7*6*5=210` possible ways. But for each group of three there are 6 ways to represent them -- ABC,ACB,BAC,BCA,CAB, and CBA. So `210/6=35`
Using combinations -- if the toppings are labeled A,B,...,G, then it does not matter whether you choose ABC,BAC, etc... Since that is the case, you can find the number of combinations of three elements chosen from seven, or `_7C_3=35`
Or you could list the possibilities with a tree diagram. With 7 options this could be tedious, and you would have to account for the repeats -- again ABC is the same as CAB.
(2) As stated, the problem is ambiguous since there is no restriction on repeats. It is conceivable that you could repeat toppings. Order a medium pizza with olives, olivs, and olives for instance. In this case the counting principle gives `7*7*7=343` ways.
(3) Also, as stated the order may matter. It might make a difference whether you put the extra cheese on top of the pepperoni and mushrooms. In this case ABC is different from BAC and there would be `7*6*5=210` or using permutations `_7P_3=210` choices.
7 x 6 x 5 = 210
3 x 2 x 1 =6
210/6 = 35