TIFR CSE 2024 | Part A | Question: 2

Question

Let $\sigma$ be a uniform random permutation of $\{1, \ldots, 100\}$. What is the probability that $\sigma(1)<\sigma(2)<\sigma(3)$ (i.e., what is the probability that the first three elements are in increasing order)?

In the options below, $n !=1 \times 2 \times \ldots \times n$.

• A. $\frac{3}{100 !}$
• B. $\frac{3 !}{100 !}$
• C. $\frac{6}{100}$
• D. $\frac{1}{6}$
• E. $\frac{1}{3}$

   

Answer 1

As given each permutation of n elements have same probability i.e,  $$\frac{1}{100!}$$

Considering a permutation =>    $6,4,7,1,2,3,.........,98,100$ 

                If we consider changing the order of first 3 elements the following are the possibilities

                                      $6,4,7,1,2,3,.........,98,100$ 

                                      $4,6,7,1,2,3,.........,98,100$ 

                                      $7,4,6,1,2,3,.........,98,100$ 

                                      $7,6,4,1,2,3,.........,98,100$ 

                                      $4,7,6,1,2,3,.........,98,100$ 

                                      $6,7,4,1,2,3,.........,98,100$ 

 

              The Valid permutation among these is   $4,6,7,1,2,3,.........,98,100$  where the

first 3 numbers  are in increasing order...So for all 6 permutations in  $100!$  as above

the valid permutation is only one…

 

The possible cases can be calculated like:     $$\frac{100!}{3!}$$

                                                    

The probability that the first three elements are in increasing order:

                                                       $\frac{\frac{100!}{3!}}{100!}$      =     $\frac{1}{6} $

 

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