The 5-digit number PQRST (all distinct digits) is such that $T\ne 0$. 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?

The correct option is 4.

Let us solve the problem step-by-step by analyzing the conditions given for the 5-digit number $PQRST$, where $P,Q,R,S,T$ are all distinct digits (i.e., numbers from 0 to 9).

Step 1: Analyze the condition for P and T
We are given that $T\ne 0$ and $P=3T$.
Since $P$ and $T$ are single digits (0 to 9) and $T\ne 0$, the possible values for $T$ and $P$ are:
- If $T=1$, then $P=3$
- If $T=2$, then $P=6$
- If $T=3$, then $P=9$

Step 2: Analyze the conditions for S, Q, and R
We are given:
1. $S$ is greater than $Q$ by 4, so $S=Q+4$.
2. $Q$ is greater than $R$ by 3, so $Q=R+3$.

Substituting $Q=R+3$ into the equation for $S$:

$S=(R+3)+4=R+7$

Since $S$ must be a valid digit (less than or equal to 9) and all digits are non-negative, let us find the possible values for $(R,Q,S)$:
- If $R=0$, then $Q=3$ and $S=7$
- If $R=1$, then $Q=4$ and $S=8$
- If $R=2$, then $Q=5$ and $S=9$
(If $R\ge 3$, then $S\ge 10$, which is not a single digit.)

Step 3: Combine cases ensuring all digits P, Q, R, S, T are distinct

Let us test the possible combinations of $(P,T)$ and $(R,Q,S)$:

Case 1: $T=1$, $P=3$
- If $(R,Q,S)=(0,3,7)$: Invalid, because $Q=3$ conflicts with $P=3$.
- If $(R,Q,S)=(1,4,8)$: Invalid, because $R=1$ conflicts with $T=1$.
- If $(R,Q,S)=(2,5,9)$: All digits $(P=3,Q=5,R=2,S=9,T=1)$ are distinct! (Valid number: 35291)

Case 2: $T=2$, $P=6$
- If $(R,Q,S)=(0,3,7)$: All digits $(P=6,Q=3,R=0,S=7,T=2)$ are distinct! (Valid number: 63072)
- If $(R,Q,S)=(1,4,8)$: All digits $(P=6,Q=4,R=1,S=8,T=2)$ are distinct! (Valid number: 64182)
- If $(R,Q,S)=(2,5,9)$: Invalid, because $R=2$ conflicts with $T=2$.

Case 3: $T=3$, $P=9$
- If $(R,Q,S)=(0,3,7)$: Invalid, because $Q=3$ conflicts with $T=3$.
- If $(R,Q,S)=(1,4,8)$: All digits $(P=9,Q=4,R=1,S=8,T=3)$ are distinct! (Valid number: 94183)
- If $(R,Q,S)=(2,5,9)$: Invalid, because $S=9$ conflicts with $P=9$.

Conclusion:
The possible 5-digit numbers are 35291, 63072, 64182, and 94183.
Thus, there are 4 such possible 5-digit numbers.