The number of seven digit positive integers formed using the digits...

The given numbers are $1, 2, 3, 4$ and sum of digits should be $12$.

Case $1: 4 3 1 1 1 1 1$ ,Here the sum of the digits is $12$.

The number of ways arranging them is $= \dfrac{7 !}{5 !} = 42$.

Case $2: 4 2 2 1 1 1 1$.

The number of ways arranging them is $= \dfrac{7 !}{2 ! 4 !} = 105$

Case $3: 3 2 2 2 1 1 1$.

The number of ways arranging them is $= \dfrac{7 !}{3 ! 3 !} = 140$

Case $4: 3 3 2 1 1 1 1$.

The number of ways arranging them is $= \dfrac{7 !}{2 ! 4 !} = 105$

Case $5: 2 2 2 2 2 1 1$

The number of ways arranging them is $= \dfrac{7 !}{5 ! 2 !} = 21$

The total number of ways are $= 42 + 105 + 140 + 105 + 21$

$= 413$

Therefore, the required answer is $413$.