<p>The total number of 3-digit numbers whose sum of digits is 10 is ________. (JEE Main 2020)</p>
Step-by-Step Solution
Key Concept: Convert the constraint to a standard form of counting non-negative integral solutions, then account for the digit constraint that each must be ≤ 9.
<p><strong>Step 1:</strong> Let the digits of 3-digit numbers be $x, y, z$ such that $x + y + z = 10$ and $x, y, z \in \{0, 1, 2, 3, \ldots, 9\}$, but $x \neq 0$ (since it's the first digit).</p><p><strong>Step 2:</strong> Let $x = t + 1$ where $t \in \{0, 1, 2, 3, \ldots, 8\}$</p><p><strong>Step 3:</strong> Substituting: $t + 1 + y + z = 10$</p><p>$\Rightarrow t + y + z = 9$</p><p><strong>Step 4:</strong> This equation has non-negative integral solutions.</p><p>Number of solutions = $\binom{9+3-1}{3-1} = \binom{11}{2} = 55$</p><p><strong>Step 5:</strong> However, we must exclude cases where $y > 9$ or $z > 9$.</p><p>When $y \geq 10$: Let $y = 10 + y'$, then $t + y' + z = -1$ (no non-negative solutions)</p><p>When $z \geq 10$: Similarly, no solutions.</p><p>But we need to subtract cases where digits exceed 9: there is 1 such case.</p><p>Total = 55 - 1 = 54</p><p>∴ Answer is (a) 54.</p>
Correct Answer: A