No of solutions for $x_1x_2x_3x_4 = 770$ Question:
Let $N$ be the set of all integral solutions of the equation $x_1x_2x_3x_4 = 770$. Find $N$
So, for this question I used cases. Using prime factorization we know that the factors are $7 , 2 ,5 ,11$.
Case 1:When $x_1,x_2,x_3,x_4$ are $7 , 11 , 5 ,2$ in all permutations then number of ways of arranging :$4!$
Case 2:When two of the numbers are already multiplied. Ex. $7,11,10,1$ and here first I have to choose two numbers and then arrange so:${4 \choose 2} \cdot 4!$
Case 3: When a pair of two numbers are multiplied. Ex. $77 , 10 , 1 , 1$ and here I have to choose two and then arrange:${4 \choose 2} \cdot \frac{4!}{2!}$
Case 4: When three numbers are multiplied. Ex.$7 , 110 , 1,1$ and here I have to choose three numbers then arrange:${4 \choose 3} \cdot \frac{4!}{2!}$
Case 5: When one number is $770$ and the others are $1$ then ways of arranging is $4$
Thus total no of ways ($N$) $= 292$. However the answer is given as $256$. Which case have I missed and is there a better method to approach this question. Any help will be appreciated!
 A: Your case 3 is not correct.  $4 \choose 2$ is the number of ways to combine two prime factors, but you count $77,10,1,1$ twice, once when you choose $7,11$ for the two and once when you choose $2,5$.  This divides the cases by $2$.  You must have added wrong, because that correction reduces the total and the answer comes out $256$ as desired.
A: There is a simpler way to solve the problem without making multiple cases.
The number $770$, as you rightly said, has $2,5,7$ and $11$ as it's prime factors. The positive integers $x_1$,$x_2$,$x_3$,$x_4$ must exist as some combination of these factors.
Let $a_1$, $a_2$,$a_3$, $a_4$ represent the exponents of 2 in each of these numbers. Similarly, assume $b_n$, $c_n$ and $d_n$ as the exponents of 5, 7 and 11 respectively. Consider the exponents of 2. These will add up as we multiply each of the $x_n$s and the sum of the exponents of 2 from each $x$ must equal the total exponent of 2 in the number 770. Hence we get the equation:
$$a_1 + a_2 + a_3 + a_4 = 1$$
Similarly for other exponents,
$$b_1 + b_2 + b_3 + b_4 = 1$$
$$c_1 + c_2 + c_3 + c_4 = 1$$
$$d_1 + d_2 + d_3 + d_4 = 1$$
The formula for the number of solutions to such an equation is quite well known, and is equal to $\binom{4+1-1}{1} = 4$
Since these are $4$ such simultaneous equations, we can easily see that the total number of solutions must be $4^4 = 256$.
In fact, you can extend this method to solve for any equation of the form $x_1.x_2.x_3....x_n = k$ as long as you know the prime factors of the number $k$.
