The number of ways in which $11^{11}$ can be written as a product of $3$ factors The number of ways in which $11^{11}$ can be written as a product of $3$ factors
Attempt:writting $11^{11} = a\times b \times c = 11^{x}\times 11^{y}\times 11^{z} = 11^{x+y+z}$
so $x+y+z = 11$ and $x,y,z\geq 0$ and $x,y,z$ all are integer
now one pair is $(2,3,6)$ similarly other pairs is $(0,0,11),(1,1,9),.....$
could some help me how to get all ordered pairs, thanks
 A: Suppose you have to select two $1$'s from horizontal string of thirteen $1$'s.
$$\underbrace{1~~~1~~~1~~~1~~~1~~~1~~~1~~~1~~~1~~~1~~~1~~~1~~~1}_{11 ~\text{times}}$$
If you have to find the number of sloutions of $x+y+z=11$, it is similar to selecting any two $1$'s in the string of $1$'s.Since it would divide all the one $1$'s into three parts (left,middle and right) giving you the value of $x,y,z$.For example ;
$$\underbrace{~1~~~1~}_{2}~~\boxed{1}~~\underbrace{ 1~~~1~~~1~~~1}_{4}~~\boxed1~~\underbrace{1~~~1~~~1~~~1~~~1}_{5}$$
Selecting $1$'s in this way gives you $x=2,y=4,z=5$.
$$\underbrace{}_{0}~~\boxed1~~~\underbrace{}_{0}~~\boxed1~~\underbrace{1~~~1~~~1~~~1~~~1~~~1~~~1~~~1~~~1~~~1~~~1}_{11}$$
Selecting $1$'s in this way gives you $x=0,y=0,z=11$.
Total number of ways of selecting two $1$'s :
$$\binom{13}{2}=78$$
It would be better if you remember a formula for solving this type of questions :
Number of ways for selecting $x_i$'s such that 
$$x_1+x_2+x_3 \dots x_r=n ~~~\text{is}~~ \binom{n+r-1}{r-1}$$
Where each $x_i \ge 0$
A: int z, y, x, sum = 0;
    for (z = 11; z >= 0; z -= 1)
        for (y = 11; y >= 0; y -= 1)
            for (x = 11; x >= 0; x -= 1)
                if (x + y + z == 11)
                    sum += 1;
    System.out.println(sum);

by a brute force algorithm i got the answer : $78$ 
for all permutation which means that (0,11,0) and (0,0,11) and (11,0,0) are considered three distinct answers, if you don't want permutation the answer is $16$ 
int z, y, x, sum = 0;
    for (z = 11; z >= 0; z -= 1)
        for (y = z; y >= 0; y -= 1)
            for (x = y; x >= 0; x -= 1)
                if (x + y + z == 11)
                    sum += 1;
    System.out.println(sum);

