Finding number of integral solutions I am really getting confused in this question.
Number of integral solutions of the equation.
$x_1x_2x_3x_4=770$
options-


*

*$2^{11}$

*$2^{10}$

*$4^4$

*$5^5$
I attemtemted it by saying that $7*11*5*2=770$
so solutions should be $4!$
then we can say $77*5*2*1=770$
so $4!$
and $77*10*1*1=770$
so $4!/2$
and then $770*1*1*1=770$
so $4!/3!$
now multiplying them i get $27648$.
i am sure i have made a big mistage here but i dont know what.
please help
 A: Case 1:  $2\cdot 5\cdot 7\cdot 11 = 770 \to 4! = 24$
Case 2: $10\cdot 7\cdot 11\cdot 1 = 770 \to 4!$
Case 3: $14\cdot 5 \cdot 11\cdot 1 = 770 \to 4!$
Case 4: $22 \cdot 5\cdot 7\cdot 1 = 770 \to 4!$
Case 5: $35 \cdot 2\cdot 11\cdot 1 = 770 \to 4!$
Case 6: $55 \cdot 2\cdot 7\cdot 1 = 770 \to 4!$
Case 7: $77\cdot 2\cdot 5\cdot 1 = 770 \to 4!$
Case 8: $70\cdot 11\cdot 1\cdot 1 = 770 \to 4\cdot 3 = 12$
Case 9: $110\cdot 7\cdot 1\cdot 1 = 770 \to 4\cdot 3 = 12$
Case 10: $154\cdot 5\cdot 1\cdot 1 = 770 \to 4\cdot 3 = 12$
Case 11: $385\cdot 2\cdot 1\cdot 1 = 770 \to 4\cdot 3 = 12$
Case 12: $770\cdot 1\cdot 1\cdot 1 = 770 \to 4$.
Thus the total number of solutions is: $7\cdot 24 + 4\cdot 12 + 4 = 220$
A: You have four buckets, $x_1$, $x_2$, $x_3$, and $x_4$, and four numbers, $2, 5, 7, 11$, to put in those buckets. As the buckets are ordered, the numbers unique, and you can put several numbers into the same bucket, there's $4^4$ ways of performing this task.
However, you're interested in inegral solution, so you ought to consider positive and negative. Let us choose signs of $x_1, \dots, x_4$. If we have chosen the signs of $x_1$, $x_2$, $x_3$, then the sign of $x_4$ is uniquely determined. There is $2\cdot 2 \cdot 2$ ways of choosing the signs of $x_1, \dots x_3$.
That is, the answer should be $4^4\cdot 2^3 = 2^{11}$.
A: 770=$7\cdot11\cdot2\cdot5$
for, $x_1x_2x_3x_4$we can place 5,7,2 and 11 all in $4 \choose 1$ ways.
So total no of solutions are $4 \choose 1$$4 \choose 1$$4 \choose 1$$4 \choose 1$=$4^4$
