The panel for a political forum consists of four Democrats, 5 republicans, 3 independents in how many orders can they be seated
2 answers:
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.
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:
Since there are 12 seats, we can seat them in 12! different ways, assuming straight line.
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:
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