Question

Four players — W, X, Y, Z — play a round-robin tournament (each plays each once).
A win gives 2 points, loss 0.
W scores more points than X.
Y wins exactly one match.
Z does not lose to X.
How many distinct possible point-tables exist for the four players?

Hint:

When dealing with round-robin constraints, isolate players with fixed win-loss counts (like “wins exactly one match”) and process forced matches first.

Answer 1

Correct Answer: 6

Step 1: Six games are played (round robin of 4), so total points \( =12 \). Y wins exactly one game, so Y's score is fixed at \( 2 \); and "Z does not lose to X" forces Z to beat X.
Step 2: Y's one win can fall against W, X, or Z, giving three skeletons for the table; each skeleton leaves only the W-X and W-Z results undecided.
Step 3: Apply "W scores more than X" to each skeleton. This keeps 1 table when Y's win is over W, 3 tables when Y's win is over X, and 2 tables when Y's win is over Z — but the table \( (W,X,Y,Z)=(4,2,2,4) \) turns up in two different skeletons, so it is not counted twice.
Step 4: The surviving, non-repeating tables are \( (4,2,2,4) \), \( (6,0,2,4) \), \( (4,0,2,6) \), \( (6,2,2,2) \).
\[ \boxed{4} \]