I'm assuming we can say each Democrat, Republican, and Independent are the same element.
Since there are 12 seats, we can seat them in 12! different ways, assuming straight line.
[tex]\text{Total(no restrictions): } 12![/tex]
However, the four democrats are the same element, since we only care about the different parties. Using this same logic, the 5 Republicans can swap without us knowing and the 3 Independents, as well.
Thus, we've overcounted by a factor of 4! 5! 3!, because each element are identical.
To counter this, we divide by the number of elements repeated:
[tex]\text{Total ways: } \frac{12!}{4!5!3!} = 27,720[/tex]
