Set A consists of 25 distinct numbers. We pick n numbers

+1.

We should understand following two things:
1. The probability of picking any n numbers from the set of 25 distinct numbers is the same. For example if we have set of numbers from 1 to 25 inclusive, then the probability we pick n=5 numbers {3,5,1,23,25} is the same as that of we pick n=5 numbers {9,10,4,6,18}. So picking any 5 numbers \(\{x_1,x_2,x_3,x_4,x_5\}\) from the set is the same.

2. Now, imagine we have chosen the set \(\{x_1,x_2,x_3,x_4,x_5\}\), where \(x_1<x_2<x_3<x_4<x_5\). We can pick this set of numbers in \(5!=120\) # of ways and only one of which, namely \(\{x_1,x_2,x_3,x_4,x_5\}\) is in ascending order. So 1 out of 120. \(P=\frac{1}{n!}=\frac{1}{5!}=\frac{1}{120}\).

According to the above the only thing we need to know is the size of the set (n) we are choosing from the initial set A.

Answer: B.