JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Permutations and Combinations

Let the number be $\overline{abc}$ with $a\in \{1,\dots ,9\}$, $b,c\in \{0,\dots ,9\}$, $a+b+c=14$.

Substitute $a^{\prime }=a-1\in \{0,\dots ,8\}$: $a^{\prime }+b+c=13$, with $0\le a^{\prime }\le 8$, $0\le b\le 9$, $0\le c\le 9$.

Number of nonnegative integer solutions with no upper bound: $\left(\genfrac{}{}{0ex}{}{13+2}{2}\right)=\left(\genfrac{}{}{0ex}{}{15}{2}\right)=105$.

Subtract solutions violating each bound (inclusion–exclusion):
$a^{\prime }\ge 9$: set $a^{\prime \prime }=a^{\prime }-9$, then $a^{\prime \prime }+b+c=4$, giving $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ solutions.
$b\ge 10$: set $b^{\prime }=b-10$, then $a^{\prime }+b^{\prime }+c=3$, giving $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ solutions.
$c\ge 10$: similarly $10$ solutions.

No two bounds can be violated simultaneously (e.g. $a^{\prime }\ge 9$ and $b\ge 10$ would need $a^{\prime \prime }+b^{\prime }+c=13-9-10=-6<0$, impossible), so no addback terms are needed.