I often solve math questions to improve my math. Today I came across an interesting permutation & combination question.
The question is as follows:
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?
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.
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.
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.