On the Saturday of a weekend softball tournament, Team A plays Team B and Team C plays Team D. Then on Sunday, the two Saturday winners play for first and second places while the two Saturday losers play for third and fourth places. There are no ties. One possible ranking of the team from first place to fourth place at the end of the tournament is the sequence ACBD. What is the total number of possible four-team ranking sequences at the end of the tournament?
+113 A and B could both be winners in the first match, and C and D could both be winners of the second match.
So really, all combinations of A, B, C and D are possible rankings here, given there are no tied matches.
So we are back to square one, in how many ways can You order the four teams A, B, C and D?
Lets look at the first possibility, that team A comes out on top, then we have these possible rankings:
A>B>C>D
A>B>D>C
A>C>B>D
A>C>D>B
A>D>B>C
A>D>C>D
This gives us 6 possible combinatioons with A as the winners. The same is true if either of the other
teams win, so we have 6 x 4 rankings = 24.
We can calculate this number by taking the faculty of 4, 4! = 4 x 3 x 2 x 1 = 24
