Permutations & Combinations
Counting three-digit numbers with digit inequality
Grade Class 12
Question:
The number of three-digit numbers $\overline{abc}$ which satisfy $a \leq b > c$ is
$2({}^9C_2+{}^9C_3+{}^9C_1)$
${}^9C_2+{}^9C_3+{}^9C_1$
$2({}^9C_2+{}^9C_3)+{}^9C_1$
None of these
Step-by-Step Solution
Key Concept: Split into exhaustive cases based on whether digits are equal or distinct, and whether $c=0$ or not.
Total $= 2{}^9C_3 + 3\cdot{}^9C_2 + {}^9C_1$, which does not match any of (A)–(C).
Correct Answer: 4