I have $8$ flavors of ice cream and $10$ types of topping.
How many options do you have if you can only take one scoop but up to $3$ toppings?
Toppings can't be repeated and the order does not matter.
Is $$8 \times 10 + 8 \times 10 \times 9 + 8 \times 10 \times 9 \times 8 + 8$$ right? I mean so you can choose what ever you like or?
Thanks a lot.
Peter
Break the problem into 2 steps. Pick the ice cream then decide on the toppings. The number of ways to do this is the number of ways to pick the ice cream times the number of ways to select the toppings. The number of ways to selecting the toppings should be partitioned into selecting 0,1,2, or 3. Find number of ways of selecting i toppings. If order doesn't matter but you can repeat, to find this count, you use a method often known as "balls in boxes" or "donut" problem solutions. You can find discussion of this type of problem in Rosen's Discrete Mathematics and Its Applications in the chapters on counting. The number of ways to select i toppings from 10 is C(9+i,i) where i=0,..,10.
If on the other hand order is not important but you cannot repeat toppings, then the answer would be 8* (C(10,0)+C(10,1)+C(10,2)+C(10,3)). If order is important and you cannot repeat it is 8*(1+10+10*9+10*9*8) ways, which is the way you were approaching it, Peter. So your answer depends whether order is important and whether toppings can be repeated.