Let E={2,4,6,8,10,12,14,16,18,20} be the set of all even numbers from 1 to 20. There are 10 even numbers.
Consider any permutation of the numbers 1,2,…,20. In this permutation, the 10 even numbers will appear in some order. For example, if we consider the numbers 2, 4, and 6, one possible relative order is (4, 2, 6), meaning 4 appears before 2, and 2 appears before 6.
We are interested in the event that 2 appears at an earlier position than any other even number. This means that among the 10 even numbers, 2 must be the first to appear in the permutation.
By symmetry, each of the 10 even numbers is equally likely to appear first among the even numbers.
There are 10! ways to arrange these 10 even numbers among themselves within the permutation.
The number of arrangements where 2 appears first is 9! (since 2 is fixed as the first, and the remaining 9 even numbers can be arranged in 9! ways).
The probability is the ratio of favorable arrangements to the total arrangements for the relative order of the even numbers:
P(2 appears first among evens)=Total number of arrangements of even numbersNumber of arrangements where 2 is first=10!9!=101