Solution: 2016 Fall Final - 18
Author: Michiel SmidQuestion
Consider a uniformly random permutation of the set $\{1,2,\dots,77\}$. Define the event
- A = "in the permutation, both 8 and 4 are to the left of 3".
(a)
1/4
(b)
None of the above.
(c)
1/3
(d)
1/2
Solution
I’m not sure what the right way to do this is
Forget the other values, let’s just focus on 3,4, and 8
There are 3! ways to arrange 3,4, and 8: $ {(3,4,8), (3,8,4), (4,3,8), (4,8,3), (8,3,4), (8,4,3) } $
Out of those 6 ways, only 2 of them have 4 and 8 to the left of 3: $ { (4,8,3), (8,4,3) } $
$ Pr(A) = \frac{2}{6} = \frac{1}{3} $