CareerCup

Total combinations = 20! / (10! * 10!)

public static void perm(char[] w,int d)
        {
            
            if (d == w.length)
                System.out.println(count++ + ". " + new String(w) );
                
            char lastSwap = ' ';
                for(int i=d; i<w.length; i++)
                {
                    if(lastSwap == w[i])
                        continue;
                    else
                        lastSwap = w[i];
                    
                    w = swap(w,d,i);
                    perm(w,d+1);
                    w = swap(w,d,i);
                }
            
        }

        public static char[] swap(char[] w, int x, int y)
        {
            char t = w[x];
            w[x] = w[y];
            w[y] = t;
            return w;
        }

A0,A1...A10 11 possibilities
B0,B1...B10 11 possibilities
altogether 121 possibilities-1=120

Identical A's and B's?

Combination of A's and B's can be a string of 2 letters or more (AB, ABB or ABBAABB). If it is a string of 5 letters then there is 2! x 2! x 2! x 2! x 2! = 2x2x2x2x2 or 2^5 combination. If 6 letter string then there is 2^6 combination and so on.

#define SIZE 21
void print(int a,int b,char str[])
{
if(a==0&&b==0)
{
str[SIZE-1]='\0';
puts(str);
return;
}
if(a!=0)
{
str[SIZE-1-a-b]='A';
print(a-1,b,str);
}
if(b!=0)
{
str[SIZE-1-a-b]='B';
print(a,b-1,str);
}
return;
}

call print(10,10,str)

public static void main(String[] args) {
        int noOfA = 3;
        int noOfB = 3;
        generateCombi(noOfA, noOfB, "");

    }

    public static void generateCombi(int noOfA, int noOfB, String s) {
        if (noOfA == 0 && noOfB == 0)
            System.out.println(s);
        if (noOfA > 0) {
            generateCombi(noOfA - 1, noOfB, s + "A");
        }

        if (noOfB > 0) {
            generateCombi(noOfA, noOfB - 1, s + "B");
        }

}

We dont need extra memory to do this.

void func(int no_A,int no_B) {

if(no_A) {
printf("A");
func(no_A-1,no_B);
}
if(no_B) {
printf("B");
func(no_A,no_B-1);
}
}

There are n choose r ways of choosing r items from n items. In this case given 20 items, he is expecting nc0+nc1+nc2+..ncn = 2^n.