The number of seven digit positive integers formed using the digits 1, 2, 3 and 4 only and sum of the digits equal to 12 is _____.

We need to count seven-digit positive integers using digits $$\{1, 2, 3, 4\}$$ with digit sum equal to 12.

Let the digits be $$d_1, d_2, \ldots, d_7$$ where each $$d_i \in \{1, 2, 3, 4\}$$ and $$\sum d_i = 12$$.

Substitute $$e_i = d_i - 1$$, so $$e_i \in \{0, 1, 2, 3\}$$ and:

$$\sum_{i=1}^{7} e_i = 12 - 7 = 5$$

We count non-negative integer solutions to $$e_1 + e_2 + \cdots + e_7 = 5$$ with $$0 \leq e_i \leq 3$$ using inclusion-exclusion.

Without upper bound: $$\binom{5+6}{6} = \binom{11}{6} = 462$$

Subtract cases where at least one $$e_i \geq 4$$: Set $$f_i = e_i - 4 \geq 0$$. The remaining variables sum to $$5 - 4 = 1$$. Number of solutions: $$\binom{1+6}{6} = 7$$. There are $$\binom{7}{1} = 7$$ choices for which variable, so we subtract $$7 \times 7 = 49$$.

Cases where two variables $$\geq 4$$: This requires sum $$\geq 8 > 5$$, which is impossible. So no correction needed.

By inclusion-exclusion:

$$462 - 49 = 413$$

The answer is $$\boxed{413}$$.