Demonstration of Role of Permutation in Probability

Let’s illustrate the power of counting techniques in the computation of probability with the following example. There are 8 students in the class namely A, B, C, D, E, F, G, H.  

B, D, F are girls &

A, C, E, G, H are boys.

What is the probability that girls occupy top three positions in the end semester exam? No two students get exactly same marks. All the students are academically equally sound.

Solution:  Total arrangements of 8 students = 8!

If we list down each individual arrangement it will be a very time consuming process.

And it will look like this.

ABCDEFGH, HABCDEFG, …………………………., ABHCDEFG

These are 8! = 40320 arrangements of 8 students.

Arrangement in which girls occupy top 3 places and boys occupy last 5 positions = 3! 5!

If we list down each individual arrangement it will be a very time consuming process.

And it will look like this.

BDFACEGH, DBFACEGH, FDBACEGH, FBDACEGH, BFDACEGH, DFBACEGH, ……….DFBAGHCE

These are 3! 5! = 720 arrangements

P( only girls occupy top 3 positions)= (3!5!)/(8!)=1/56

We see that in the long run if these 8 students take 56 exams then girls are in the top 3 positions in only one exam.

Listing down 40320 outcomes of the total cases and identifying 720 outcomes of favorable cases is a time consuming and tedious process. Use of permutation here simplifies everything. This example can be related to other similar situations like what is the probability the three girls are first to die when considering the mortality experience of these people. We assume that they have the same health status.