I solved it with slightly different method 
The idea is to imagine the array with all 3's removed. There would be blocks of 1's and 2's. Since a[1] = 1 and a[n] = 1, the array would look like this: 11..1 \; 22..2 \; 11..1 \; 22..2 \; 11..1. Clearly no. of blocks will be odd with alternating 1's and 2's. Now imagine adding 3's. To “fix” a block of size s you need to use 3 exactly s-1 times. For example, to fix 11111 you need to use 3 four times (to get 131313131).
So, if there are k blocks, you require at least x+y-k threes to fix, Now there are additional (n-x-y ) - (x+y-k) threes left, there are also k-1 borders between 1's and 2's in which we have to insert these additional 3's. So, we gotta select (n+k-2x-2y) spaces among total of k-1 spaces. There are \binom{k-1}{n+k-2x-2y} ways to do so!
Now, to form blocks, note that total size of blocks of 1's is exactly x, let w_{i} be the length of the i^{th} block of 1's. Also, there are ceil(k/2) blocks of 1's and floor(k/2) blocks of 2's We need to find positive integral solutions of -
w_{1} + w_{2}...+w_{ceil(k/2)} = x.
This is simply, \binom{x-1}{ceil(k/2)-1}. Similar result for no. of ways of making blocks of 2's is \binom{y-1}{floor(k/2)-1}.
So finally the answer would be to sum \binom{x-1}{ceil(k/2)-1}\binom{y-1}{floor(k/2)-1}\binom{k-1}{n+k-2x-2y} for all odd k from 1 to n.
Link for my submission: CodeChef: Practical coding for everyone