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Since order matters, use permutations. You will need to do two seperate things since it is a group of 3 and then a group of 5, and multiply them together \(\Large \frac{8!}{(8-3)!}\times\frac{8!}{(8-5)!}\)
Wait, I think I did that wrong...
Oh, sorry I didn't fully read the question. You only want to do the people in the three chairs, so just do \(\Large\frac{8!}{(8-3)!}\)
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1.6 x 2.6
Remember that 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
Oh so C will be my Official Answer
No, not C. REmeber the (8-5)! part
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\(\Large\frac{8!}{(8-3)!}\rightarrow \frac{8*7*6*5*4*3*2*1}{5!}\rightarrow \frac{8*7*6*5*4*3*2*1}{5*4*3*2*1}\)
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