Doubt: GATE 2007-24

Question

https://gateoverflow.in/1222/gate2007-24

Following is the question from GATE-2017-24

Suppose we uniformly and randomly select a permutation from the 20! permutations of 1,2,3…,20. What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?

1) In the above question what could be different equally likely outcomes?

2)The obvious one is all the 52! permutations.

Another is 10 even numbers (2,4…20) possible [Scroll down for Arjun sir's answer]

Why should the above be considered as equally likely outcomes?

3) What could be other ways to approach the problem apart from these two?

 

Lets begin with the 2nd question, then 3rd which will result to the 1st question's answer.

Comments

check my method ( check 2 answers with comments, https://gateoverflow.in/240811/topological-sort )

total 20 numbers, 10 odd and 10 even, just as topological sequences 2 -->4 , 2 -->6 ,....2 --->20

keeping all even numbers as one unit.

total permutations = $\frac{20\; ! }{ (10(even\; numbers\; are\; one\; unit)\; !\; *\; 1(odd\; numbers\; are\; unique)\; !)} \;*\;9(2\; is \;fixed \;at\; first\; place)\;! $ = $\frac{20\; ! }{10}$

Probability = $\frac{(\frac{20\; ! }{10})}{20\;!}$ = $\frac{1 }{10}$

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@Shaik Masthan

I did not understand what "keeping all even numbers as one unit." mean?

Is this what you meant?

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@Shaik Masthan

Ignore my previous comment I understood what you did.