Basic statistics question

Factual Questions
1

I was about to choose some characters for a computer game when I began to wonder : how many combinations (order isn’t important, I mean [a,b,c,d] is the same combination than [c,a,d,b]) of 4 different elements (characters, in this case) in a set of 14 are there? What formula should I apply? I learnt that at some point in the past, but I can’t figure it out now…

2

14!/(4!)(10!)=1001 different combinations

3

Thanks a lot!
But could someone explain me (roughly) the logic behind this formula? Not that I really need to know ( I didn’t need to know the formula, either), but I’m curious…I suppose 14! is the total number of possible combinations of 5 different ordered elements (don’t know wheter “ordered” is the correct word in english or not…I hope younunderstand what I mean), but why should I divide it by 4! times 10! when I’m not concerned about order?
So, if someone has too much time in his/her hands, and doesn’t mind answering…

4

Your OP asks for how many different 4 number sets you can get from a set of size 14. Your follow up mentions 5, not 4. If you want 5 then it would be:

14!/(5!)(9!)=2002 different combinations.

I don’t want to get into the ‘philosophy’ of Combinations but the formula is:

nCr=n!/((r!)(n-r)!)

If order is not important (permutations) then the formula is:

nPr=n!/(n-r)!

5

Oops, permutation=if order is important

6

14! is the number of ways to arrange 14 different objects.

Imagine seeing all 14 objects in a row. You only care about the first 4 of them. The arrangement of the last 10 doesn’t matter.
There are 10! ways to arrange them. So you divide by 10! to get 14!/10!

This is now the number of ways to take 4 objects from 14 where you care about the order.

There are 4! ways to arrange 4 objects, so now you divide by 4! to get 14!/ (10! * 4!)

This is now the number of ways to take 4 objects from 14 where you don’t care about the order.

I hope this helps.

Ed

7

Also, you need to divide by r! for combinations because, if you have a 4 member set, there are 4! combinations where they have the same members but different order. That’s why, if order is not important, that you divide by r!.

8

Thanks to you both…It’s perfectly clear, now!

I’ll be able to play my computer game with my mind in peace (at least as long as I won’t wonder about the likehood of my character being killed by a +2 arrow while wearing an armor class 5 studded leather :wink: )