The problem requires us to determine the correct gym schedule for seven people (P, Q, R, X, T, A, K) across seven days of the week (Sunday to Saturday) based on a set of clues, and then identify the incorrect combination among the given options.

Analyzing the Clues and Constraints

Let's break down the information provided:

• People: P, Q, R, X, T, A, K
• Days: Sunday (1), Monday (2), Tuesday (3), Wednesday (4), Thursday (5), Friday (6), Saturday (7)
• Clue 1: R goes to the gym on the weekend (Sunday or Saturday).
• Clue 2: K goes after X, and K goes before Q. This implies the order is X ... K ... Q in terms of days.
• Clue 3 & 5: The number of people before A equals the number of people after Q. Also, Q goes after A. Let $D_A$ be the day number for A and $D_Q$ be the day number for Q.
  • Number before A = $D_A - 1$
  • Number after Q = $7 - D_Q$
  • From Clue 3: $D_A - 1 = 7 - D_Q \implies D_A + D_Q = 8$.
  • From Clue 5: $D_Q > D_A$.
  Possible (A, Q) pairs satisfying $D_A + D_Q = 8$ and $D_Q > D_A$:
  • If $D_A=1$ (Sun), $D_Q=7$ (Sat)
  • If $D_A=2$ (Mon), $D_Q=6$ (Fri)
  • If $D_A=3$ (Tue), $D_Q=5$ (Thu)
• Clue 4: K and X do not go to the gym on Wednesday (Day 4).
• Clue 6: Only one person goes to the gym between R and X. This means their day numbers differ by 2, i.e., $|D_R - D_X| = 2$.
• Clue 7: P likes to go to the gym on weekdays (Monday to Friday).

Step-by-Step Deduction

Step 1: Determine possible positions for R and X using Clues 1 and 6.

R must be on Sunday (1) or Saturday (7).

• Case 1: R is on Sunday (1). Using $|D_R - D_X| = 2$, we get $|1 - D_X| = 2$. This means $D_X$ must be 3 (Tuesday). This is valid as X is not on Wednesday.
• Case 2: R is on Saturday (7). Using $|D_R - D_X| = 2$, we get $|7 - D_X| = 2$. This means $D_X$ must be 5 (Thursday). This is valid as X is not on Wednesday.

So, the possible pairs for (R, X) are (Sun, Tue) or (Sat, Thu).

Step 2: Evaluate the possible (A, Q) pairs with the (R, X) placements.

Possible (A, Q) pairs are: (A=1, Q=7), (A=2, Q=6), (A=3, Q=5).

Evaluating Case 1 (R=Sun(1), X=Tue(3)):

• Subcase 1a: Test (A=1, Q=7). This conflicts because R is already assigned to Sunday (1). Invalid.
• Subcase 1b: Test (A=2, Q=6). So, A = Monday (2), Q = Friday (6).
  • We have R=Sun(1), A=Mon(2), X=Tue(3), Q=Fri(6).
  • Now use Clue 2 (X ... K ... Q): X is Tue(3), Q is Fri(6). K must be between Tuesday and Friday. Possible days for K are Wednesday (4) or Thursday (5).
  • Use Clue 4 (K not on Wednesday): K cannot be on Wednesday (4). Therefore, K must be on Thursday (5).
  • Current assignment: R=Sun(1), A=Mon(2), X=Tue(3), K=Thu(5), Q=Fri(6).
  • Days remaining: Wednesday (4) and Saturday (7). People remaining: P and T.
  • Use Clue 7 (P on weekdays): P must be on a weekday. The remaining weekdays are Wednesday (4). So, P = Wed(4).
  • The last person T must be on the remaining day, Saturday (7).
  • Final Schedule Check: Sun(R), Mon(A), Tue(X), Wed(P), Thu(K), Fri(Q), Sat(T).
    • R weekend: Yes (Sun).
    • X..K..Q order: Tue..Thu..Fri. Yes.
    • Before A = After Q: $D_A=2$, $D_Q=6$. Before A = 1 (Sun). After Q = 1 (Sat). Yes.
    • K,X not Wed: K=Thu, X=Tue. Yes.
    • Q after A: $D_Q=6 > D_A=2$. Yes.
    • 1 between R & X: R=Sun(1), X=Tue(3). Mon(2) is between. Yes.
    • P weekdays: P=Wed(4). Yes.
    This schedule is valid.
• Subcase 1c: Test (A=3, Q=5). So, A = Tuesday (3), Q = Thursday (5). This conflicts because X is already assigned to Tuesday (3). Invalid.

Evaluating Case 2 (R=Sat(7), X=Thu(5)):

• Subcase 2a: Test (A=1, Q=7). A = Sunday (1), Q = Saturday (7). This conflicts because R is already assigned to Saturday (7). Invalid.
• Subcase 2b: Test (A=2, Q=6). A = Monday (2), Q = Friday (6).
  • We have R=Sat(7), X=Thu(5), A=Mon(2), Q=Fri(6).
  • Use Clue 2 (X ... K ... Q): X is Thu(5), Q is Fri(6). K must be between Thursday and Friday. There is no day between Thursday and Friday. Impossible to place K. Invalid.
• Subcase 2c: Test (A=3, Q=5). A = Tuesday (3), Q = Thursday (5). This conflicts because X is already assigned to Thursday (5). Invalid.

Since Case 2 yields no valid schedules, the only possible schedule is the one derived from Case 1b.

Final Schedule and Incorrect Combination

The only valid schedule determined is:

Now, let's check the given options against this schedule:

• 1. A - Monday: This combination is present in the schedule. Correct.
• 2. Q - Saturday: This combination is NOT present in the schedule (Q is on Friday). Incorrect.
• 3. K - Thursday: This combination is present in the schedule. Correct.
• 4. R - Sunday: This combination is present in the schedule. Correct.
• 5. Q - Friday: This combination is present in the schedule. Correct.

The question asks for the incorrect combination. Based on our derived schedule, the combination 'Q - Saturday' is incorrect.