How many three-digit numbers can be formed using digits {$1$, $6$, $9$, $9$, $9$,}? If I had $5$ distinct digits $(1,2,3,4,5)$ I would do it like so:

$\frac{5!}{(5-3)!} = 60$

But I don't understand what to do if I have $3$ repeating digits

*

*First Number - $999$ 

*Second Number (Take all distinct digits) = $3$!

*Third Number (Take $2$ $9$'s) = $C(1,1) * C(3,2) * C(2,1)= 3$!


Total Numbers = $1 + 6 + 6 = 13$

$C(1,1)$ - How many repeating numbers are there ? Only $9$ is repeating and choose to include only that .
$C(3,2)$ - How many $9$'s are there ? How many are we using ?
$C(2,1)$ - After including $9$, only one place left . How many ways to fill that ?
