Probability that 2 appears at an earlier position than any other even number in a permutation of 1-20 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?
My approach : 
2 in first position ==> 1*19! (1 is the position of 2) 
2 in second position ==> 10 *1 *18! 
2 in third position ==> 10*9 *1 *17! 
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.
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2 in 11th position ==> 10*9*8*7*6*5*4*3*2*1*1*9!
But this was asked as a multiple choice question in GATE 2007 and the options were 
(A) 1/2   (B) 1/10  (C) 9!/20!  (D) None of these
I am not able to reduce this analysis to a given option. 
 A: All even numbers have the same probability of being first. And there are $10$ of them. So the probability is $\dfrac{1}{10}$.
A: Let $\Omega$ be the sample space which discribes all possible permutations
of the $20$ numbers. Clearly, $|\Omega|=20!$ and we let each permutation be
equally likely. To find all permutations where $2$ appears before all elements
of the set $E=\{k\in \mathbb{N} \mid 4\le k\le 20 \land k  \text{ even} \} $.
The idea now is to consider all possible positions where the elements of
$E \cup \{2\} $ could appear and for each possible positioning of those
elements there are $9!$ permutations of them where 2 comes before all of them.
The remaining $10$ elements can be permuted in $10!$ ways, so we obtain
\begin{align*}
 P(\text{"2 appears before all elements of }S\text{"})
 =\frac{\displaystyle \binom{20}{10} \cdot 9!\cdot 10!}{|\Omega|}
 =  \frac{\displaystyle \frac{20!}{10!\cdot 10!}\cdot 9!\cdot 10!}{20!}
 = \frac{9!}{10!}=\frac{1}{10}
.\end{align*}
A: The odd numbers do not matter here. The probability 2 comes before the other 9 evens is
$$\dfrac{(\text{# of ways to pick 2})(\text{# of ways to pick remaining evens})}{(\text{# of ways to order 10 evens})} = \dfrac{1\cdot 9!}{10!} = \dfrac{1}{10}$$
A: I took the trouble of calculating the long expression (1*19!+10*1*18!+10*9*1*17!+10*9*8*1*16!+..........+10*9*8*7*6*5*4*3*2*1*1*9!)/20!
and it is = 1/10.
Stay happy!
Although David's approach is elegant, I just wanted to be absolutely sure about it
A: Well my approach goes like this:
As question is asked about 2 comes before any other even no., only restriction is imposed on 2 not on any other no.
So considering when 2 is at first place:
To keep 2 at first place=1 way
and rest 19 numbers can be arranged in 19! ways
Since only condition is to be satisfied is that 2 comes from other even numbers
So,permutation of numbers 2,4,6,8,...........is different from 2,4,8,6........
as permutation means set of different ordering. Therefore, when 2 is at first place total no. of ways rest numbers can be arranged is 19!
Now, when 2 is at second place
Any odd number can take place at first position, so total no. of ways are 10 now 2 is at second lace, so only way is desired to hold up then rest 9 odd numbers and even numbers other 2 will be following and these numbers can be arranged in 18!
so total no. of ways=10*1*18!
When 2 is at3rd position
No. of ways=10*9*1*17!
(No repetition is allowed since permutation means different ordering)
When 2 is at 4th position 
No. of ways=10*9*8*1*16!
Similarly all odd numbers can be placed before 2
and maximum position up-to which 2 can be placed is 11th
Any odd numbers can take places from 1 to 10
Again, the fact is to be considered since permutation demands different ordering therefore ordering of different set of numbers is counted as one single permutation
So, total no. of ways when 2 is at 11th position is:
10*9*8*7*6*5*4*3*2*1*1*9! 
so total no. of ways when 2 can be shifted holding the condition all even number comes after 2 are:
1*19!+10*1*18!+10*9*1*17!+10*9*8*1*16!+............+10*9*8*7*6*5*4*3*2*1*1*9!
Total no ways in which 20 numbers can be permuted are 20!
So, required probability:(1*19!+10*1*18!+10*9*1*17!+10*9*8*1*16!+..........+10*9*8*7*6*5*4*3*2*1*1*9!)/20!
