There are nine 'different' slots we can fill in for each value. Thus, we get 9! different ways, including repetition.
Now, we need to account for repetition. This can be visualised as separate letters as below:
s₁ u c₁ c₂ e₁ s₂ s₃ e₂ s₄
We can arrange them like this, and we'd never know the difference:
s₁ u c₂ c₁ e₂ s₃ s₂ e₁ s₄
s₄ u c₁ c₂ e₁ s₁ s₃ e₂ s₂
Thus, we have overcounted by a factor of 2! (c's can change in 2! ways), 2! (e's can change in 2! ways), and 4! (s' can change in 4! ways).
Therefore, our final number of permutations becomes:
