Mike D. answered • 05/03/21
Tutor
4.9
(154)
Experienced high school discrete math teacher
Well lets say the easiest is E, and the hardest is H
Then if x are the other 8 questions we can place the E and H as follows
EHxxxxxxxx
xEHxxxxxxx
xxEHxxxxxx
xxxEHxxxxx
xxxxEHxxxx
xxxxxEHxxx
xxxxxxEHxx
xxxxxxxEHx
xxxxxxxxEH
Obviously you can switch the E, H here.
So there are 9 x 2 = 18 ways of placing the EH or HE.
Once these are placed we have 8 spots to fill with 8 questions. Order is important to first can be placed in 8 ways, second in 7 ... so 8 x 7 x 6 x ... x 1 = 8 ! ways
So final answer = 18 x 8! = 725760 ways
Mike D.
There are 10! ways of placing the letters and 1/5 of the letters are E or H so it seems that 10!/5 also gives the correct answer, but I can't easily justify that logically.
Report
05/03/21
Ally J.
Thank you, but would there also be another way of doing this? Maybe a little easier?05/03/21