My resource comes from https://faculty.math.illinois.edu/~hildebr/408/408combinatorial.pdf.

Part a)
The notes linked above says that a) is a number of unordered samples of size r, without replacement, from a set of n objects. In english, which means, let's say you have 7 ingredients that go on a sandwich and you can only choose 4 out of those 7, also, you can't double/triple/quadruple on one ingredient. The order of ingredients that go on a sandwich doesn't matter as long as they are on there.

For example, you have 1) turkey, 2) salami, 3) american cheese, 4) pepperjack, 5) onions, 6) spinach and 7) olives. Then that number from a) gives you how many different sandwiches you can make with the rules I mentioned above; you can't choose the same ingredient twice or more (comes from "without replacement") and the order of the ingredients doesn't matter ("unordered samples"). Size r refers to the number of ingredients allowed to choose, in this case is 4. And a set of n objects are the total number of ingredients allowed to choose from, which is 7.

turkey, turkey, salami, pepperjack is not allowed; "without replacement"
turkey, salami, pepperjack, onions = salami, turkey, onions, pepperjack; "unordered"

Part b}
Well, I approached this problem by trying to make b} into a "correct" format, like the one you see from part a). Then you could write the whole thing as two combinations multiplied together, (10 2) * (8 5). The explanation would be the same for each one. You can think of each combination as a sandwich shop. Like (10 2) sandwich shop has 10 ingredients with 2 choices and (8 5) sandwich shop has 8 ingredients with 5 choices. But if you multiply them together, it would be the total number of possible sandwich combinations from these two shops.

Let me know what you think

This post was edited by liljohn_jy on Dec 10 2020 06:30pm

Thank you! I think I perfectly understand for part a), correct me if I'm wrong. My understanding is that (n r)=n!/(r!(n-r)!) which basically means that one item cannot be chosen more than once, and the order doesn't matter. So if you would pick the same ingredients but in different order, it would actually be the same thing.
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For part b} I'm not quite sure sure how you went from formula b} to (10 2) * (8 5). Could you please expand on that?

This post was edited by haxe on Dec 10 2020 07:51pm

Yes, you got part a).

So by definition (n r) = n! / (r! (n-r)!) = (n(n - 1) ... (n - r +1)) / r!. You can also find this in Formulas part on the link I provided.

10! / (5! * 3! * 2!) = (10 * 9 * 8!) / (5! * (8 - 5)! * 2!); Rewriting the terms
(10 * 9 * 8!) / (5! * (8 - 5)! * 2!) = ((10 * 9) / 2!) * (8! /(5! * (8 - 5)!)); Breaking it up to two combinations

Looking at the terms individually
1st term: (10*9)/2! = (10 2); this is the (n r) = (n(n-1)...(n-r+1)) / r! part. This case n = 10 and r = 2.
2nd term: (8! / (5! * (8 - 5)!)) = (8 5); this is the (n r) = n! / (r! (n-r)!) part. This case n = 8 and r = 5.

So if you substitute,

10! / (5! * 3! * 2!) = (10 2) * (8 5)


I know it's a bit hairy but try to follow along. I think i didn't make any stupid mistakes along. Let me know

This post was edited by liljohn_jy on Dec 10 2020 09:02pm

Makes sense to me now! thank you very much sir, really appreciated it!

nicce, thanks for the gold