# Basic statistics question **URL:** \ **Category:** Factual Questions\ **Created:** [July 25, 2002, 8:42pm UTC](https://boards.straightdope.com/t/basic-statistics-question/120826 "2002-07-25T20:42:01Z")\ **Posts on this page:** 8\ **Page:** 1
**Author:** ![clairobscur](https://avatars.discourse-cdn.com/v4/letter/c/839c29/32.png) [@clairobscur](https://boards.straightdope.com/u/clairobscur)\ **Post date:** [July 25, 2002, 8:42pm UTC](https://boards.straightdope.com/t/basic-statistics-question/120826/1 "2002-07-25T20:42:01Z")
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… ---
**Author:** ![BlinkingDuck](https://avatars.discourse-cdn.com/v4/letter/b/eada6e/32.png) [@BlinkingDuck](https://boards.straightdope.com/u/BlinkingDuck)\ **Post date:** [July 25, 2002, 8:46pm UTC](https://boards.straightdope.com/t/basic-statistics-question/120826/2 "2002-07-25T20:46:13Z")
14!/(4!)(10!)=1001 different combinations ---
**Author:** ![clairobscur](https://avatars.discourse-cdn.com/v4/letter/c/839c29/32.png) [@clairobscur](https://boards.straightdope.com/u/clairobscur)\ **Post date:** [July 25, 2002, 9:14pm UTC](https://boards.straightdope.com/t/basic-statistics-question/120826/3 "2002-07-25T21:14:45Z")
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… ---
**Author:** ![BlinkingDuck](https://avatars.discourse-cdn.com/v4/letter/b/eada6e/32.png) [@BlinkingDuck](https://boards.straightdope.com/u/BlinkingDuck)\ **Post date:** [July 25, 2002, 9:43pm UTC](https://boards.straightdope.com/t/basic-statistics-question/120826/4 "2002-07-25T21:43:00Z")
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)! ---
**Author:** ![BlinkingDuck](https://avatars.discourse-cdn.com/v4/letter/b/eada6e/32.png) [@BlinkingDuck](https://boards.straightdope.com/u/BlinkingDuck)\ **Post date:** [July 25, 2002, 9:44pm UTC](https://boards.straightdope.com/t/basic-statistics-question/120826/5 "2002-07-25T21:44:57Z")
Oops, permutation=if order is important ---
**Author:** ![suranyi](https://avatars.discourse-cdn.com/v4/letter/s/e36b37/32.png) [@suranyi](https://boards.straightdope.com/u/suranyi)\ **Post date:** [July 25, 2002, 9:47pm UTC](https://boards.straightdope.com/t/basic-statistics-question/120826/6 "2002-07-25T21:47:48Z")
> [@](#): > > \*Originally posted by clairobscur \* > \*\*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… \*\* 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 ---
**Author:** ![BlinkingDuck](https://avatars.discourse-cdn.com/v4/letter/b/eada6e/32.png) [@BlinkingDuck](https://boards.straightdope.com/u/BlinkingDuck)\ **Post date:** [July 25, 2002, 9:50pm UTC](https://boards.straightdope.com/t/basic-statistics-question/120826/7 "2002-07-25T21:50:46Z")
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!. ---
**Author:** ![clairobscur](https://avatars.discourse-cdn.com/v4/letter/c/839c29/32.png) [@clairobscur](https://boards.straightdope.com/u/clairobscur)\ **Post date:** [July 25, 2002, 10:35pm UTC](https://boards.straightdope.com/t/basic-statistics-question/120826/8 "2002-07-25T22:35:37Z")
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 😉 )