This problem requires us to determine the correct combination of a student and their admission day based on a set of logical conditions. We need to arrange seven students (J, K, L, M, N, O, P) into seven days of the week (Monday to Sunday).

Step-by-Step Deduction of the Schedule

Let's break down the information and build the schedule step-by-step:

• Given: Seven students (J, K, L, M, N, O, P) and seven days (Monday to Sunday).
• Clue 1: J takes admission on Wednesday.
  • We can fix J's position: Wed: J
• Clue 2: Only one student takes admission between J and M.
  • Since J is on Wednesday, M can be on Monday (with Tuesday in between) or Friday (with Thursday in between).
  • If M is Monday: Mon(M), Tue(?), Wed(J). This is possible.
  • If M is Friday: Wed(J), Thu(?), Fri(M). This is also possible.
• Clue 3: L takes admission before M and after K.
  • This establishes a sequence: K ... L ... M.
  • If M were Monday (from Clue 2), L would have to be admitted *before* Monday, which is impossible.
  • Therefore, M cannot be Monday. M must be on Friday.
  • So, the sequence is J (Wed), Thu (?), M (Fri).
  • Now, K ... L ... M (Fri). L must be before Friday, and K must be before L. L can be Tuesday or Thursday. K must be before L.
• Clue 4: Two students take admission between K and N. K takes admission before N.
  • This means there are exactly two days between K's admission day and N's admission day, with K coming first. If K is on day $d_K$, N is on day $d_N$, then $d_N = d_K + 3$.
• Combining Clues for K, L, M:
  • We know: Wed(J), Thu(?), Fri(M).
  • Sequence: K ... L ... M(Fri).
  • Possible slots for L: Monday, Tuesday, Thursday.
  • Possible slots for K: Monday, Tuesday.
  • If L is on Thursday: K must be Monday or Tuesday.
    • If K is Monday: Mon(K), Tue(?), Wed(J), Thu(L), Fri(M).
    • Check Clue 4: Two between K (Mon) and N ($N = Mon + 3 = Thu$). But L is on Thursday. This combination is impossible.
  • If L is on Tuesday: K must be on Monday.
    • Arrangement: Mon(K), Tue(L), Wed(J), Thu(?), Fri(M).
    • Check Clue 4: Two between K (Mon) and N ($N = Mon + 3 = Thu$). This fits perfectly.
    • Updated schedule: Mon(K), Tue(L), Wed(J), Thu(N), Fri(M).
• Remaining Students and Days:
  • Students left: O, P.
  • Days left: Saturday, Sunday.
  • Current Schedule: Mon(K), Tue(L), Wed(J), Thu(N), Fri(M).
• Clue 5: O takes admission on one of the days after M.
  • M is on Friday. O can be on Saturday or Sunday.
• Clue 6: P does not take admission on Sunday.
  • If O takes Saturday, P must take Sunday. But P cannot take Sunday.
  • Therefore, O must take Sunday, and P must take Saturday.

Final Schedule

Based on the deductions, the complete schedule is:

Evaluating the Options

Now let's check the given options against our derived schedule:

1. K – Monday: This matches our schedule. Correct.
2. L – Sunday: Our schedule shows L admitted on Tuesday. Incorrect.
3. M – Saturday: Our schedule shows M admitted on Friday. Incorrect.
4. N – Friday: Our schedule shows N admitted on Thursday. Incorrect.
5. O – Tuesday: Our schedule shows O admitted on Sunday. Incorrect.

The only correct combination is K – Monday.