The number of natural numbers, between 212 and 999 , such that the sum of their digits is 15 , is

Let the digits be $$a, b, c$$ , where  $$a \ge 2$$, and $$b, c \ge 0$$, and all ≤ 9.

We want $$ a + b + c = 15 $$ with the number between 212 and 999.

Let, $$ a = 2$$

If $$a = 2$$ , then the number must be ≥ 212, so $$b \ge 1$$

And, $$ b + c = 13$$

Valid pairs are $$ (4,9),(5,8),(6,7),(7,6),(8,5),(9,4) $$

That’s 6 pairs, all satisfy $$ b \ge 1$$ So, 6 numbers are obtained in this case. 

Similarly, we fix for other values of a.

a = 3, b+c = 12

Possible choices: (3,9) to (9,3) → 7 numbers

a = 4 , b+c = 11

Possible choices: (2,9) to (9,2) → 8 numbers

a = 5 , b+c = 10

Possible choices: (1,9) to (9,1) → 9 numbers

a = 6 , b+c = 9

Possible choices: (0,9) to (9,0) → 10 numbers

a = 7 , b+c = 8

Possible choices: (0,8) to (8,0) → 9 numbers

a = 8 , b+c = 7

Possible choices: (0,7) to (7,0) → 8 numbers

a = 9 , b+c = 6

Possible choices: (0,6) to (6,0) → 7 numbers

So, the total possible choices of numbers: 6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64