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

Each player plays 3 matches. A win gives 2 points, so possible scores are 0, 2, 4, 6. 
We must enumerate outcomes under constraints.
Step 1: Process Y’s condition. 
Y wins exactly one match. 
Thus Y earns exactly \(2\) points. 
Y must defeat exactly one of W, X, Z.
Step 2: Process Z’s condition. 
“Z does not lose to X” means: 
Z either beats X or draws with X (draw is impossible), 
so Z must beat X. 
Thus: \[ Z \to X \] Step 3: Consider each of Y’s possible wins. 
Y beats one of the three: W, X, or Z. 
We check all scenarios, making sure W scores more than X.
Case 1: Y beats W. 
Then Y loses to X and Z. Y=2 points. 
Z already beats X. 
We enumerate all remaining games: \[ W\text{ vs }X,\quad W\text{ vs }Z,\quad X\text{ vs }Y(\text{X wins}),\quad Z\text{ vs }Y(\text{Z wins}) \] Checking all valid assignments where W>X yields 2 valid point-tables. 
Case 2: Y beats X. 
Then Y loses to W and Z. X already loses to Z and Y, so X has at most 2 points. Enumerating all remaining games while keeping W>X gives 3 valid point-tables. 
Case 3: Y beats Z. 
Then Y loses to W and X. 
But Z must beat X. 
We enumerate remaining matches: \[ W\text{ vs }X,\quad W\text{ vs }Z \] Only arrangements where W>X survive. 
This case gives 1 valid point-table.
Step 4: Add all valid point-tables. 
\[ 2 + 3 + 1 = 6 \] Final Answer: \(\boxed{6}\)

Answer 2

Method: case-split on Y's single win.
Six matches are played in all (each pair meets once), so 12 points are shared out in total (2 per match, no draws). Since Y wins exactly one of the three matches Y plays, Y's own score is fixed at \( 2 \). Also, since Z does not lose to X, Z must beat X in their match.

Y's single win can only be against W, against X, or against Z — three cases. In each case the other two Y-matches are automatically losses for Y, which pins down two more match results. That leaves only the W-vs-X and W-vs-Z matches free (4 combinations each), and imposing "W's score is more than X's score" prunes them down:

Case Y beats W (so X beats Y, Z beats Y): only W beating both X and Z survives the test, giving the table \( W=4, X=2, Y=2, Z=4 \).
Case Y beats X (so W beats Y, Z beats Y): three of the four W-X/W-Z combinations survive, giving tables \( (6,0,2,4) \), \( (4,0,2,6) \), \( (4,2,2,4) \).
Case Y beats Z (so W beats Y, X beats Y): two combinations survive, giving \( (6,2,2,2) \) and \( (4,2,2,4) \).

Collecting the distinct score-tuples \( (W,X,Y,Z) \) across all three cases and removing the repeat of \( (4,2,2,4) \) that shows up twice, exactly four different point tables remain.

\[ \boxed{4} \]