Get inspired by the success stories of our students in IIT JAM 2021. Learn More

For Students who are passionate for Mathematics and want to pursue it for higher studies in India and abroad.

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

numbers?

- Algorithmic Sense
- Combination and Permutation Skills

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),

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.

- 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.

What to do to shape your Career in Mathematics after 12th?

From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:

- What are some of the best colleges for Mathematics that you can aim to apply for after high school?
- How can you strategically opt for less known colleges and prepare yourself for the best universities in India or Abroad for your Masters or Ph.D. Programs?
- What are the best universities for MS, MMath, and Ph.D. Programs in India?
- What topics in Mathematics are really needed to crack some great Masters or Ph.D. level entrances?
- How can you pursue a Ph.D. in Mathematics outside India?
- What are the 5 ways Cheenta can help you to pursue Higher Mathematics in India and abroad?

Cheenta has taken an initiative of helping College and High School Passout Students with its "Open Seminars" and "Open for all Math Camps". These events are extremely useful for students who are really passionate for Mathematic and want to pursue their career in it.

Cheenta is a knowledge partner of Aditya Birla Education Academy

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

JOIN TRIAL
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.

Yeah, you are right. I will make the changes.

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)

Yeah, you are right. I will make the changes.