My logic for the second example was as follows: we have four options for ice cream flavor so we start with a 4. Then, we have five options for topping 1 because we can either have no topping, topping 1, topping 2, topping 3, or topping 4. For the following topping, we have four options because we have all the previous options minus whichever one we already used. So on and so forth. So when I was solving I thought the answer would be 4 x 5 x 4 x 3 x 2
Once I heard the explanation of the right method, it made sense to me because I do realize now that my method wouldn’t make sense for cases in which there are no toppings (or only 1 or 2 toppings) in that you can have multiple slots filled by the option for none.
My question is, how do I identify which kinds of questions require this binary yes/no type of logic (topping vs no topping) versus the logic from the first example where we are considering that if you have already used a certain letter, then this affects the number of possibilities for the next letter and so on? I guess what I’m trying to ask is, why in the second example do we not need to care about the number of topping options or which ones have already been used?
Thank you so much for your reply, I appreciate the time you’ve taken. One quick follow up question:
I think the second question you asked me about underlying assumption (Can we have more than one topping for an ice cream concoction? If so, what are the cases involved?) may have been the root of my misunderstanding. When I attempted the original question, I thought it mattered which toppings we chose. For the following, let’s assume that we only have one flavor option, not 4, to simplify the number of cases. If my 4 topping options are raspberries, strawberries, marshmallows, and chocolate sauce, I had assumed that it would count differently to have raspberries and strawberries (2 toppings) than to have marshmallows and chocolate (2 toppings). But if I understood the solution correctly, we were supposed to think about 2 toppings as one case only. i.e. there are five cases: 1 topping, 2 toppings, 3 toppings, 4 toppings, no toppings. But I was thinking about the question as having more cases: 4 cases for one topping (choosing chocolate sauce x= choosing strawberries), 12 cases for two toppings (because strawberries+raspberries is different than strawberries and marshmallows), 4 cases for 3 toppings, 1 case for 4 toppings, and 1 case for no toppings.
Indeed. Assuming u can’t chain flavors together in a single ice cream concoction, then each flavored ice-cream concoction allows for any of the 5 cases you mentioned.
Now to work how many possibilities of each case you can have:
0 toppings from 4 possible toppings → {\binom{4} {0}}
1 toppings from 4 possible toppings → {\binom{4} {1}}
2 toppings from 4 possible toppings → {\binom{4} {2}}
3 toppings from 4 possible toppings → {\binom{4} {3}}
4 toppings from 4 possible toppings → {\binom{4} {4}}
Your final answer is then just 4 \left({\binom{4}{0}} + {\binom{4}{1}} + {\binom{4}{2}} + {\binom{4} {3}} + {\binom{4}{4}}\right) = 4 \cdot 2^4 = 64, where the outer 4 represents the number of choices available when choosing a particular flavor.
As a bonus, we used the below fact to write the sum of all those combinations in a more accessible format:
{\binom{n}{0}} + {\binom{n}{1}} + \ldots + {\binom{n}{n}} = 2^n, and why this holds follows directly from binomial theorem when for x = y = 1.
This idea looks exactly like what i wrote out above except you have 6 cases instead of 12 for two toppings. (Strawberries, Raspberries) isn’t any different from (Raspberries, Strawberries) in context, right? Other than that, I think your logic matches what i wrote above and so is a correct way to go about this problem.
