# The number of seven digit integers whose sum of the digits is ten, formed by using the digits 1, 2 and 3 only is

$\begin{array}{1 1} 55 \\ 66 \\ 77 \\ 88 \end{array}$

The seven digits whose sum of the digits $=10$ using $1,2,3$ as their digits may be
$1111123$ or $1111222$
No. of $7$ digit numbers using $1111123=\large\frac{7!}{5!}=$$42 No. of 7 digit numbers using 1111222= \large\frac{7!}{4!.3!}=$$35$
$\therefore$ the required total number of integers$=42+35=77$