The number of natural numbers, between $212$ and $999$, such that the sum of their digits is $15$, is \rule{2cm}{0.4pt}.
Step-by-Step Solution
Key Concept: Fix the hundreds digit $a\in\{2,3,\dots,9\}$. For each, count solutions of $b+c=15-a$ with $0\le b,c\le 9$. For $a=2$ exclude pairs giving numbers $\le 212$.
Count $3$-digit numbers $abc$ with $a+b+c=15,\,1\le a\le 9,\,0\le b,c\le 9.$
Stars-and-bars with inclusion-exclusion: substitute $a'=a-1$ so $a'+b+c=14$ with $0\le a'\le 8,\,0\le b,c\le 9.$
$\binom{16}{2}-\binom{7}{2}-2\binom{6}{2}=120-21-30=69.$
Exclude numbers $\le 212$ with digit sum $15$:
\quad $a=1$: $b+c=14$, pairs $(5,9),(6,8),(7,7),(8,6),(9,5)$ — $5$ numbers.
\quad $a=2,\ 200\le abc\le 212$: $b+c=13$ with $(b,c)\in\{(0,13),(1,12)\}$ — none valid.
So between $212$ and $999$: $69-5=64.$
Correct Answer: 64