Question

In how many orders can four broken scanners and four broken computers be repaired if:
a- there are no restrictions ?
b-the scanners must be repaired first?
c-the computers must be repaired first?
d-if the order of repairs has no restrictions and is done at random, what is the probability that a scanner is repaired first?

Answer 1

a- First you have 8 machines to choose, then 7, then 6, etc. so it's \[8 \cdot 7 \cdot 6 \dotsm \cdot 1 = 8! = 40320.\]b- You have four choices, then 3, then 2, then 1 and you run out of scanners. Then you have 4 choices again, then 3, etc. so it's\[(4 \cdot 3 \cdot 2 \cdot 1) \cdot (4\cdot 3 \cdot 2 \cdot 1) = 4! \cdot 4! = (4!)^2 = 576.\]c- Same as above.
d- For the first choice you have 8 machines and 4 scanners, so it's\[\frac{4}{8} = \frac{1}{2}.\]

Answer 2

Thanks :)