The answer is 56.
Let's analyze why the answer is 56. We're looking for 6-digit numbers where the sum of the digits equals 51. Since each digit can be at most 9, we can approach this as a combinatorial problem with constraints.
The minimum possible sum for a 6-digit number is 1 (first digit) + 0 + 0 + 0 + 0 + 0 = 1. The maximum possible sum is 9 + 9 + 9 + 9 + 9 + 9 = 54. Since we want the digits to sum to 51, this is a feasible problem.
Let the 6-digit number be represented as $d_1 d_2 d_3 d_4 d_5 d_6$. We have the condition:
$d_1 + d_2 + d_3 + d_4 + d_5 + d_6 = 51$
where $1 \le d_1 \le 9$ and $0 \le d_2, d_3, d_4, d_5, d_6 \le 9$.
To simplify the problem, let $d_1' = d_1 - 1$. Then $0 \le d_1' \le 8$, and our equation becomes:
$d_1' + 1 + d_2 + d_3 + d_4 + d_5 + d_6 = 51$
$d_1' + d_2 + d_3 + d_4 + d_5 + d_6 = 50$
Now, we have $0 \le d_1' \le 8$ and $0 \le d_2, d_3, d_4, d_5, d_6 \le 9$. Without the upper bound constraints, the number of non-negative integer solutions to $d_1' + d_2 + d_3 + d_4 + d_5 + d_6 = 50$ is given by $\binom{50 + 6 - 1}{6 - 1} = \binom{55}{5}$. However, this includes solutions where some $d_i > 9$. We need to subtract those invalid cases.
It is easier to calculate number of solutions with $d_1, d_2, d_3, d_4, d_5, d_6 \leq 9$. Let $x_i = 9 - d_i$, so $x_i \geq 0$ and $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 6 9 - 51 = 54 - 51 = 3$. Then, using stars and bars, there are $\binom{3+6-1}{6-1} = \binom{8}{5} = \binom{8}{3} = \frac{8 7 6}{3 2 * 1} = 56$.
Therefore, there are 56 such numbers.
