JEE Main 2025 (Online) 28th January Evening ShiftQuestion 1
The number of natural numbers, between $212$ and $999$, such that the sum of their digits is $15$, is $\_\_\_\_\_\_\_\_$ .
Permutation and Combination
Answer$64$
Solution
Concept. Fix the hundreds digit $a$, then count digit pairs $(b,c)$ with $b+c=15-a$. Formulas used. (1) Digits $b,c\in\{0..9\}$ with $b+c=s$: $s+1$ ways if $s\le9$, $19-s$ ways if $s\ge9$Step 1 - $a=3$ to $9$. $s=15-a=12,11,10,9,8,7,6$, so counts $7,8,9,10,9,8,7$. Sum $=58$. Step 2 - $a=2$. $b+c=13$: $249,258,267,276,285,294$, all $>212$. So $6$. Step 3 - Total. $58+6=64$. Answer: 64