JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Permutations and Combinations

A three-digit number $xyz$ has $x\in \{1,\dots ,9\}$, $y,z\in \{0,\dots ,9\}$. Numbers with $x=1$ and digit sum 15 satisfy $y+z=14$; the resulting numbers 159, 168, 177, 186, 195 are all less than 212 and so fall outside the given range.

For $x=2$ to $9$, count pairs $(y,z)$ with $0\le y,z\le 9$ and $y+z=15-x$:

$x=2$: $y+z=13$ → 6 pairs
$x=3$: $y+z=12$ → 7 pairs
$x=4$: $y+z=11$ → 8 pairs
$x=5$: $y+z=10$ → 9 pairs
$x=6$: $y+z=9$ → 10 pairs
$x=7$: $y+z=8$ → 9 pairs
$x=8$: $y+z=7$ → 8 pairs
$x=9$: $y+z=6$ → 7 pairs

Total $=6+7+8+9+10+9+8+7=64$.