# Permutation & Combination Problem (Math) **URL:** \ **Category:** Factual Questions\ **Created:** [May 2, 2014, 2:03am UTC](https://boards.straightdope.com/t/permutation-combination-problem-math/687480 "2014-05-02T02:03:40Z")\ **Posts on this page:** 8\ **Page:** 1
**Author:** ![MaverocK](https://avatars.discourse-cdn.com/v4/letter/m/9de0a6/32.png) [@MaverocK](https://boards.straightdope.com/u/MaverocK)\ **Post date:** [May 2, 2014, 2:03am UTC](https://boards.straightdope.com/t/permutation-combination-problem-math/687480/1 "2014-05-02T02:03:40Z")
I often solve math questions to improve my math. Today I came across an interesting permutation & combination question. The question is as follows: > [@](#): > > 6 people (named A, B, C, D, E, F) are in a line in a supermarket. > In how many possible arrangements is C between A and B? > > Options: > A. 30 > B. 60 > C. 120 > **D. 240 (\<- Correct answer)** > E. 360 The previous question was like: “In how many possible arrangements C is in front of A”. My answer was: 6!/2 which is 360. Because in one half, C is in front of A, and in the other half A is in front of C. (No calculation required, common sense & interpretation & mathematical intuition is sufficient to find the correct answer I believe.) But this one really puzzled me. I will be grateful if you provide a clear explanation. Is there any way to solve this question without writing down all combinations? ---
**Author:** ![Topologist](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/topologist/32/3208_2.png) [@Topologist](https://boards.straightdope.com/u/Topologist)\ **Post date:** [May 2, 2014, 2:26am UTC](https://boards.straightdope.com/t/permutation-combination-problem-math/687480/2 "2014-05-02T02:26:47Z")
You can solve it very similarly to how you solved the other problem. In any of the 6! particular orderings, the letters A, B, and C appear in 3 of the positions in some particular order. There are 3! different possible orders for A, B, and C, in 2 of which C appears between A and B. So, in 2/3! = 1/3 of the orderings, the letters are in the correct order. Therefore, there are 6!/3 orderings that satisfy the condition. ---
**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\ **Post date:** [May 2, 2014, 2:42am UTC](https://boards.straightdope.com/t/permutation-combination-problem-math/687480/3 "2014-05-02T02:42:55Z")
> [@Topologist](#): > > You can solve it very similarly to how you solved the other problem. In any of the 6! particular orderings, the letters A, B, and C appear in 3 of the positions in some particular order. There are 3! different possible orders for A, B, and C, in 2 of which C appears between A and B. So, in 2/3! = 1/3 of the orderings, the letters are in the correct order. Therefore, there are 6!/3 orderings that satisfy the condition. Yeah, I had to think about it for a minute, and then I was struck by this insight after noticing that the correct answer was 6!/3 and thinking about the OP’s solution to the first problem. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [May 2, 2014, 3:02am UTC](https://boards.straightdope.com/t/permutation-combination-problem-math/687480/4 "2014-05-02T03:02:53Z")
Indeed, even without having to count that there are 2 allowed out of 3! potential orderings of {A, B, C}, one could slightly more elegantly simply note that there is 1 valid out of 3 potential choices for which of {A, B, C} goes in the middle. ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [May 3, 2014, 1:10pm UTC](https://boards.straightdope.com/t/permutation-combination-problem-math/687480/5 "2014-05-03T13:10:00Z")
I assume you mean “somewhere between” not “immediately between”? Ie. if the former, ADCEFB is valid, but not the latter case? ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [May 3, 2014, 10:31pm UTC](https://boards.straightdope.com/t/permutation-combination-problem-math/687480/6 "2014-05-03T22:31:59Z")
By symmetry the number of case that C is between A and B is the same as the number in which A is between B and C and the same as the number in which B is between A and C. These three possibilities are mutually exclusive and exhaust the 6! = 720 cases, so each must happen 240 times. ---
**Author:** ![KarlGauss](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/karlgauss/32/3713_2.png) [@KarlGauss](https://boards.straightdope.com/u/KarlGauss)\ **Post date:** [May 3, 2014, 11:34pm UTC](https://boards.straightdope.com/t/permutation-combination-problem-math/687480/7 "2014-05-03T23:34:02Z")
> [@Hari\_Seldon](#): > > By symmetry the number of case that C is between A and B is the same as the number in which A is between B and C and the same as the number in which B is between A and C. These three possibilities are mutually exclusive and exhaust the 6! = 720 cases, so each must happen 240 times. Very elegant. Lovely. ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [May 5, 2014, 11:02pm UTC](https://boards.straightdope.com/t/permutation-combination-problem-math/687480/8 "2014-05-05T23:02:39Z")
> [@KarlGauss](#): > > Very elegant. Lovely. Why thank you!