我无意中发现了我的这个老答案，我必须承认，我有点困惑，我给了这样奇怪的代码。
下面是一个更好的版本：

#include <stdio.h>

int CalculateNumberOfPaths(int x, int y)
{
    int numberOfPaths = 0;
    
    // If you're arrived
    if (x == 0 && y == 0) {
        return 1;
    }
    
    // If you're out of bound
    if (x < 0 || y < 0) {
        return 0;
    }
    
    // Trying all possible left move (x axis)
    for (int i = 1; i <= x; ++i) {
        numberOfPaths += CalculateNumberOfPaths(x - i, y);
    }

    // Trying all possible down move (y axis)
    for (int j = 1; j <= y; ++j) {
        numberOfPaths += CalculateNumberOfPaths(x, y - j);
    }
    
    return numberOfPaths;
}

int main(void)
{
    printf("The number of paths Alice and her friends have is : %d", CalculateNumberOfPaths(4, 4));
    
    return 0;
}

如果需要限制最大轴移动，只需在for循环中添加&& i <= 2或其他移动限制。

最好的赌注我可以有与描述您的问题：

#include <stdio.h>

int numOfPaths(int x, int y)
{   
    // If you're arrived
    if (x == 0 && y == 0) {
        return 1;
    }
    // If you're out of bound
    if (x < 0 || y < 0) {
        return 0;
    }

    // Trying the 4 possibility
    return numOfPaths(x - 1, y) + numOfPaths(x, y - 1) + numOfPaths(x - 2, y) + numOfPaths(x, y - 2);
}

int main(void)
{
    printf("The number of paths Alice and her friends have is : %d", numOfPaths(2, 1));
    
    return 0;
}