billzhao wrote:First, the total number of arrangements is 7!
Among the arrangments, the number of arrangements in which Carmelo defeats LeBron and Carmelo defeats Lebron must be the same as there are no ties.
So the number of arrangements of finishes in which Carmelo defeats LeBron must be 7!/2; similarly, the number of arrangements of finishes in which LeBron defeats Carmelo must also be 7!/2.
I hope the explanation is clear.
This is not the best explanation you provided for this solution. 7!/2! Can be explained this way: If the seven slots are arranged thus:
1 2 3 4 5 6 7 , where 1 means the first place and 2 means 2nd and so on, then it is clear from the problem that Lebron can never be in 1. Now for all (C, L) pairs, L can never come before C. Now in permutation whenever there are restrictions involving pairs of things, we use 2! to isolated these cases. So this explains the short method above, since there are 7! ways of arranging all of them.
For a simpler example, say that A B C are to seated in a row but A can never come before B.
A B C
B A C
C B A
C A B
B C A
A C B
If we implemented this restriction we will be left with,
B A C
C B A
B C A
We got here by 3!/2! = 3. All cases involving A coming before B (Counting from the left) have been eliminated by diving by 2!.
