Alice randomly selects a permutation of #(1, 2,\cdots\ \cdots\ , 74, 75).# What is the probability that she selects a permutation where #75# precedes 7?# ##" "## # (Note: In the permutation #[1, 4, 5, 6, 3, 2], 4# precedes #2#.)

1 Answer

#1/2#

Explanation:

I'm going to assume that with the permutation we're taking, we're permuting all of the numbers 1 through 75.

The denominator of this probability fraction is the number of ways we can permute 75 numbers, which is #75!#

So how many ways can we permute the numbers such that 7 is before 75?

To answer this, let's look at the population 7, 75. We can permute this with half the time 75 preceding 7.

Or what about the population 1, 7, 75. There are 6 ways to permute these 3 numbers and half of them have 75 preceding 7:

75, 1, 7
75, 7, 1
1, 75, 7

Or what about the population of 1, 2, 7, 75, with 24 ways to permute the numbers? There are 12 ways to have 75 precede 7:

75, 1, 2, 7
75, 1, 7, 2
75, 2, 1, 7
75, 2, 7, 1
75, 7, 1, 2
75, 7, 2, 1

1, 75, 2, 7
1, 75, 7, 2
2, 75, 1, 7
2, 75, 7, 1

1, 2, 75, 7
2, 1, 75, 7

And so by observation we can see that 75 will precede 7 half the time.