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

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

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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$=42+35=77$
answered Aug 9, 2013 by rvidyagovindarajan_1

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