Number of Seven Digit Integers with Sum of Digits Equal to Ten

To find the number of seven-digit integers whose digits sum to $10$ and are formed using only the digits $1, 2,$ and $3$, let the seven digits be denoted by $d_1, d_2, d_3, d_4, d_5, d_6, d_7$, where $d_i \in \{1, 2, 3\}$ for each $i = 1, 2, \dots, 7$.

We are given the condition:
$d_1 + d_2 + d_3 + d_4 + d_5 + d_6 + d_7 = 10$

Let $x_i = d_i - 1$. Since $d_i \in \{1, 2, 3\}$, we have $x_i \in \{0, 1, 2\}$.

Substituting $d_i = x_i + 1$ into the sum:
$(x_1 + 1) + (x_2 + 1) + \dots + (x_7 + 1) = 10$
$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 3$

We now need to find the number of non-negative integer solutions to:
$x_1 + x_2 + \dots + x_7 = 3 \quad \text{such that} \quad 0 \le x_i \le 2 \text{ for each } i$

We can analyze the possible partitions of the sum $3$:

1. Partition $1 + 1 + 1$:
   Three of the $x_i$'s are equal to $1$, and the remaining four are $0$.
   This corresponds to the digits being three $2$'s and four $1$'s:
   $\text{Number of arrangements} = \frac{7!}{3! \, 4!} = \frac{5040}{6 \times 24} = 35$

2. Partition $2 + 1 + 0$:
   One of the $x_i$'s is equal to $2$, one is equal to $1$, and the remaining five are $0$.
   This corresponds to the digits being one $3$, one $2$, and five $1$'s:
   $\text{Number of arrangements} = \frac{7!}{1! \, 1! \, 5!} = \frac{5040}{120} = 42$

3. Partition $3 + 0 + 0$:
   One of the $x_i$'s is equal to $3$, which corresponds to a digit $d_i = 4$. Since the only permitted digits are $1, 2,$ and $3$, this case is not allowed.

Adding the valid arrangements together:
$\text{Total number of seven-digit integers} = 35 + 42 = 77$

Therefore, the correct option is C.