A probability problem by Yeldo Pailo

Supone that each digit is a box with balls inside (the number of balls in each box will be its digit)

At least, all the boxes have a ball $\Rightarrow$ three balls have to be spread in the seven boxes. This can be done of two ways:

$1) (°°,°)$ (two balls in one box and other apart)

ways of select where are two balls together $=\binom{7}{1}$

ways of select where is the other ball $=\binom{6}{1}$

$=7\times6=\boxed{42}$

$2)(°,°,°)$ (three balls in separate box) $=\binom{7}{3}=\boxed{35}$

Number of seven digits, with sum equal to $10$ formed using the $1,2$ and $3$ only is $42+35=\boxed{77}$