I get how to solve this problem by counting the possibilities, but IDK how to do it without actually needing to count the possibilities.
I think it would be 7+6+5+4+3+2+1 as the number of times you can pick 2 things out of 8
I think it would be 6+5+4+3+2+1 as the number of times you can pick 3 things out of 5 but that doesn't sit right with me, it seems too high.
The number of ways to create a set of three items from five is 10. Your instinct was right.
Let's start with simple examples. We have a set of three distinguishable things. How many different sets of one of them can we have? 3. A or B or C. How many different sets of two of them can we have? 3. A, B or A,C or B, C. But what about B,A for example? We are not interested in order. A,B and B,A have the same elements and so are the same set. How many different sets of three can we have? 1, obviously.
OK let's try a set of two drawn from a set of eight distinguishable items. So A, B, C, D, E, F, G, and H are eight distinguishable items. We could have A, B or A, C or A, D, or A, E or A, F or A, G or A, H or B, C, or B, D, or B, E or B, F, or B, G or B, H or C, D or C, E or C, F or C, G or C, H or D, E or D, F or D, G or D, H or E, F or E, G, or E, H or F, G or F, H or G, H. That is 28 in total.
Now if if you want you could try four from ten distinguishable items through that method. As the numbers get bigger, this method of total enumeration becomes time consuming and very error prone. It is easier to remember a formula.
We want to know how many distinct ways we can draw k items from a set of n distinguishable items.
The answer is \(\displaystyle \dbinom{n}{k} = \dfrac{n!}{k! * (n - k)!}.\)
We define \(\displaystyle 0! = 1 \text { and } m \in \mathbb N^+ \implies m! = m * (m - 1)!.\)
So 0! is 1, 1! is 1, 2! is 2, 3! is 6, 4! is 24, and so on.
Let's use that formula on our examples.
One from three.
\(\displaystyle \dbinom{3}{1} = \dfrac{3!}{1! * (3 - 1)!} = \dfrac{3 * 2!}{1 * 2!} = 3.\)
That agrees with our method of total enumeration.
Two from three.
\(\displaystyle \dbinom{3}{2} = \dfrac{3!}{2! * (3 - 2)!} = \dfrac{3 * 2!}{2! * 1!} = \dfrac{3}{1} = 3.\)
That agrees. How about three from three?
\(\displaystyle \dbinom{3}{3} = \dfrac{3!}{3! * (3 - 3)!} = \dfrac{1}{0!} = \dfrac{1}{1} = 1.\)
That agrees. How about two from eight?
\(\displaystyle \dbinom{8}{2} = \dfrac{8!}{2! * (8 - 2)!} = \dfrac{8 * 7!}{2! * 6!} = \dfrac{8 * 7 * 6!}{2! * 6!} = \dfrac{8 * 7}{2} = 28.\)
Seems like a pretty neat formula.