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

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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$=42+35=77$

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