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If you are allowed to use extra memory, you can improve the solution by caching calculated value so that the second recursion branch becomes more efficient.
Also, pascals(x, y) == pascals(x, x-y) once you calculate one of these, set the other one in the cache as well.

Here is O(x) solution

def pascal(x, y):
    assert x <= y
    x = min(x, y - x)
    result = 1
    for i in xrange(x):
        result = (result * (y - i)) / (i + 1)
    return result

You can calculate values of pascals triangle using combinatorics

Yth (zero-indexed) value at Xth level can be calculated as (X-1) choose Y

top down dynamic programming.

static int getPascalValue(int i, int j, HashMap<String, Integer> map, int[] cnt) {
        //base cases: left col or diagonal
        if( j == 0 || i == j)
            return 1;
        if(map.containsKey(i + " " + j))
            return map.get(i + " " + j);

        int res =  getPascalValue(i -1,j, map, cnt) + getPascalValue(i-1,j-1, map,cnt);
        map.put(i + " " + j, res);
        cnt[0]++;
        return res;
    }

    public static void main(String[] args) {
        int cnt[] = new int[1];
        HashMap<String, Integer> map = new HashMap<>();
        out.println(getPascalValue(40, 20, map, cnt) + " cnt: " + cnt[0]);
    }

407575348 cnt: 400

public static int pascal(int x, int y){
        if(y == 0 || x == y) return 1;
        else return pascal(x -1 , y - 1) + pascal(x - 1, y);

}