Q. The 5-digit number PQRST (all distinct digits) is such that T ≠ 0. P is thrice T. S is greater than Q by 4, while Q is greater than R by 3. How many such 5-digit numbers are possible?

Explanation : 

1. The 5‐digit number is P Q R S T with all digits distinct and T ≠ 0.
2. Given:
• P = 3·T
• Q = R + 3
• S = Q + 4

3. Since P=3T must be a single digit (0–9) and T≠0, the only possibilities for T are 1, 2 or 3:
– T=1 ⇒ P=3
– T=2 ⇒ P=6
– T=3 ⇒ P=9

4. Also Q=R+3 ≤9 ⇒ R≤6, and S=Q+4=R+7 ≤9 ⇒ R≤2. Hence R∈{0,1,2}.

5. Check each T and each R∈{0,1,2}, discarding any repeat digits:

Case A: T=1, P=3
R=0 ⇒ Q=3 (conflicts with P=3) ✗
R=2 ⇒ Q=5, S=9 → digits {3,5,2,9,1} all distinct ✓
→ Number = 35291

Case B: T=2, P=6
R=0 ⇒ Q=3, S=7 → {6,3,0,7,2} distinct ✓ → 63072
R=1 ⇒ Q=4, S=8 → {6,4,1,8,2} distinct ✓ → 64182
R=2 ⇒ Q=5, S=9 → conflicts with T=2 or repeats none but R=2=T? actually R=2 conflicts with T=2 ✗

Case C: T=3, P=9
R=0 ⇒ Q=3 conflicts with T=3 ✗
R=1 ⇒ Q=4, S=8 → {9,4,1,8,3} distinct ✓ → 94183
R=2 ⇒ Q=5, S=9 conflicts with P=9 ✗

6. Valid numbers found: 35291, 63072, 64182, 94183 → total 4.

Answer: (b) Only 4 such numbers.