INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

May 4, 2020

Let's Permute | ISI MStat 2018 PSB Problem 3

This problem is an easy application of the basic algorithmic ideas to approach a combinatorics problem using permutation and combination and basic counting principles. Enjoy this problem 3 from ISI MStat 2018 PSB.


Consider all permutations of the integers \(1,2, \ldots, 100\) . In how many
of these permutations will the \(25^{th}\) number be the minimum of the
first 25 numbers and the \(50^{th}\) number be the minimum of the first 50



Whenever you are counting, think in terms of algorithms or steps in which you will do the task given in the problem and count each of the steps. Now the conjunction in between the steps ( and \(\cap\) , or \(\cup\) ) will show you the way.


Select the first 50 numbers. Fix the minimum element. Out of the 49 other elements, select the first 25 numbers. Fix the minimum element of these 25 numbers. Then permute the rest.

AND for each of these selections of two sets (Multiplication Principle),

Step 2

Once you select the three sets, the minimum of each of these sets is already fixed. So, for each of these sets, you have to permute the other first 24 numbers (1 - 24) and the next 24 numbers (from 26 - 49) and the other 50 numbers (50 - 100).

You can do that in \(24! \times 24! \times 50! \) ways.

Therefore, you can do the whole task in \( {100 \choose 50} \times { 49 \choose 25 } \times 24! \times 24! \times 50!\) ways.

Food for Thought

  • Do the problem for \(2n\) numbers.
  • Do the problem if you want to fix the minimum \(k\) numbers in each set out of \(2n\) numbers.
  • Do the problem if you want to fix the minimum \(n\) numbers in each set out of \(2n\) numbers. Does it match your expectation?
  • Generalize and Degeneralize and make your own problem out of this. Comment it below. We will update it here if we think the problem is appropriate.

4 comments on “Let's Permute | ISI MStat 2018 PSB Problem 3”

  1. Shouldn't it be choosing 50 numbers from100 in 100C50 ways, and these 50 numbers can be ordered in 1 way, so say the first 50 numbers are arranged in decreasing order, now fixing the position of the 25th and 50th number the first 24 numbers can be arranged in 24! Ways, and the numbers between 25th and 50th can also be arranged in 24! , So it becomes 100C50*24!*24! ways total number of possibilities.

  2. I feel there is a snag somewhere. Let P be the random permutation.See that there are no restrictions in choosing P(51),..P(100) .The 50th element would have to be the minimum of the remaining 50. Again there are no restrictions in choosing P(49),..,P(26). The 25th element should be the minimum among the rest and then again no restrictions in choosing P(24),..P(1). So, the total number of ways=100*99*...*51*49*..26*24*..*1
    =100! /(25*50)

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.