<p>A 5-digit number is formed without repetition using the digits 0, 1, 2, 3, 4. The probability that the number is divisible by 4 is</p>
Step-by-Step Solution
Key Concept: Divisibility by 4 depends only on the last two digits. Enumerate valid last-two-digit pairs, then count valid completions.
<p>Total 5-digit numbers (first digit ≠ 0): $4 \times 4! = 96$.</p><p>For divisibility by 4, last two digits (tens, units) must form a number divisible by 4. Valid pairs from $\{0,1,2,3,4\}$ (distinct):</p><p>(0,4), (1,2), (2,0), (2,4), (3,2), (4,0) — 6 pairs.</p><p>Count valid 5-digit completions for each pair (first digit ≠ 0):</p><ul><li>(0,4), (2,0), (4,0): remaining digits all non-zero → $3! = 6$ each</li><li>(1,2), (2,4), (3,2): 0 in remaining → $3! - 2! = 4$ each</li></ul><p>Favorable $= 3 \times 6 + 3 \times 4 = 30$.</p><p>$P = \dfrac{30}{96} = \dfrac{5}{16}$</p>
Correct Answer: A