Number of ways where 2 occurs before any even number Question:-
Suppose we have a set of numbers from $ 1,2,3...20.$ What are the number of permutations where $2$ occurs at an earlier position than any other even number?
My approach :-
As $2$ should appear before any other even number, I clubbed all other even numbers into one set $B$ , $2$ into a set $A$ and remaining $10$ odd numbers as it is.
Now number of ways where A should occur before B can be calculated as 
$(12!)/(2!)$.
As within B, 9 even numbers can also be permuted, the total number of ways are:-
$(12! * 9!) /(2!)$.
Now if they would have asked the probability that 2 appears at an earlier position than any other even number, I think the answer should be:-
$(12! * 9!)/(20! * 2!)$ .
What is wrong with my approach ? 
 A: By clubbing together all even numbers other than $2$ into $B$, you rule out those permutations where the odd and even numbers are interspersed, e.g. 2,4,3,6,... is not counted in your approach.
And as to the actual answer:  it should be exactly 1 in 10 by symmetry! That is, in every permutation one of the 10 even numbers will be the first one, and by symmetry the probability of that being $2$ is the same as for any other even number.
A: An approach similar to yours is to group the odd numbers into an ordered set with $10!$ ways of ordering them and the even numbers into another ordered set with $9!$ ways of ordering them (the 2 must appear first). Then in order to make your permutation you either choose the first number not already chosen from the odd numbers or choose the first number not already chosen from the even numbers and continue this until you have chosen all 20 numbers. As you have to choose 10 of each there are $\binom{20}{10}$ ways of doing this. Therefore the total number of permutations with 2 the first of the even numbers is:
$$10!\cdot9!\cdot\binom{20}{10} = \frac{20!}{10}$$
A: There are $20!$ possible permutations. Looking only at the even numbers, there are ten possible positions where the two may appear and only one of them is allowed. Each of these positions occurs the same number of times, namely
$$20! \cdot \frac1{10}$$
A: Maybe split on the occurrence of 2. This can be at position 1 to 11 (the last when all odd one are before it).
If it is at $i$ we have $(i-1)$ positions before it where only odd numbers can appear and $20-i$ after it where all other evens and remaining odds can appear.
So we have $\binom{10}{i-1}$ ways to pick the odd ones before $2$, and then $(i-1)!$ ways of ordering them. Then at $i$ we put 2, and then we have $20-i$ numbers left for as many places. 
So I get $$\sum_{i=1}^{11} \binom{10}{i-1} (i-1)!(20-i)!$$ ways in total.
There is probably a nicer way...
