<s>[INST] How many eight-digit numbers can be formed with the numbers 2, 2, 2, 3, 3, 3, 3, 4, 4? I know how to determine the problem when we form nine-digit numbers that is $\frac{9!}{(3!\cdot 4!\cdot 2!)}$. But what about eight-digit numbers?
 [/INST] We need to omit exactly one of the numbers.  So any $8$-digit number induces the multiset of numbers:
\begin{align*}
& \{2,2,3,3,3,3,4,4\}, \\
& \{2,2,2,3,3,3,4,4\}, \text{or } \\
& \{2,2,2,3,3,3,3,4\}.
\end{align*}
These give rise to
\begin{align*}
\binom{8}{2,4,2}+\binom{8}{3,3,2}+\binom{8}{3,4,1} &= 420+560+280\\
&= 1260
\end{align*}
possible $8$-digit numbers.
</s>