JEE Main 2024 — Permutation Combination Question with Solution

$\mathrm{N}=\mathrm{ab}\mathrm{c}$ (i) $\begin{aligned} & \text { All distinct digits } \\ & \mathrm{a}+\mathrm{b}+\mathrm{c}=14 \\ & \mathrm{a} \geq 1 \\ & \mathrm{~b}, \mathrm{c} \in\{0 \text { to } 9\} \end{aligned}$ $\begin{array}{lll} by hit \& trial : & 8 cases \\ (6,5,3) & (8,6,0) & (9,4,1) \\ (7,6,1) & (8,5,1) & (9,3,2) \\ (7,5,2) & (8,4,2) & \\ (7,4,3) & (9,5,0) & \end{array}$ (ii) 2 same, 1 diff $a=b$;c $2\mathrm{a}+\mathrm{c}=14$ by values : $\begin{aligned} & \left.\begin{array}{l} (3,8) \\ (4,6) \\ (5,4) \\ (6,2) \\ (7,0) \end{array}\right\} \frac{3 !}{2 !} \times 5-1 \\ & =14 \text { cases } \end{aligned}$ (iii) all same : $\begin{aligned} & \begin{array}{l} 3 a=14 \\ a=\frac{14}{3} \times \text { rejected } \\ 0 \text { cases } \end{array} \end{aligned}$ Hence, Total cases: $\begin{aligned} & 8 \times 3 !+2 \times(4)+14 \\ & =48+22 \\ & =70 \end{aligned}$