Question

The number of $7$-digit numbers formed by the digits $1,$ $2$ and $3$ only whose sum of the digits equals $10,$ is

• A. $42$
• B. $55$
• C. $77$
• D. $35$

Answer 1

The correct option is C $77$

Let $1$ appear $x$ times, $2$ appear $y$ times and $3$ appear $z$ times.
Then, $x+y+z=7$ and $x+2y+3z=10$
Solving above equations, we get
$y+2z=3$
Possible values are
$y=1,z=1$ and $x=5$
or $y=3,z=0$ and $x=4$
$\therefore$ Total count $=\frac{7!}{5!1!1!}+\frac{7!}{4!0!3!}=77$