Question:

Step 1: Analyze the constraint on the first digit from condition (i). Since $4000 \leq N < 6000$, the first digit $a$ must be either 4 or 5. $$a \in \{4, 5\}$$ This gives us **2 choices** for $a$. Step 2: Analyze the constraint on the last digit from condition (ii). Since $N$ must be a multiple of 5, the last digit $d$ must be either 0 or 5. $$d \in \{0, 5\}$$ This gives us **2 choices** for $d$. Step 3: Analyze the constraint on the middle digits from condition (iii). We need to find all pairs $(b, c)$ such that $3 \leq b < c \leq 6$. The possible values for $b$ and $c$ are from the set $\{3, 4, 5, 6\}$. Let us enumerate all valid pairs where $b < c$: - When $b = 3$: $(3,4), (3,5), (3,6)$ — 3 pairs - When $b = 4$: $(4,5), (4,6)$ — 2 pairs - When $b = 5$: $(5,6)$ — 1 pair Total number of valid pairs: $3 + 2 + 1 = 6$ pairs. Step 4: Apply the multiplication principle to find the total count. Since the choices for $a$, the pair $(b,c)$, and $d$ are independent, we multiply the number of choices: $$\text{Total} = (\text{choices for } a) \times (\text{choices for pairs } (b,c)) \times (\text{choices for } d)$$ $$\text{Total} = 2 \times 6 \times 2 = 24$$ The number of 4-digit numbers satisfying all three conditions is **24**. The answer is **Option 3: 24**.