Given, $x_1.x_2.x_3.x_4=770=\mathrm{2.5.7.11}$

Let positive integer $x_1=2^{a_1}{.5}^{b_1}.7^{c_1}{.11}^{d_1}$

Let positive integer $x_2=2^{a_2}{.5}^{b_2}{.7}^{c_2}{.11}^{d_2}$

Let positive integer $x_3=2^{a_3}{.5}^{b_3}{.7}^{c_3}{.11}^{d_3}$

Let positive integer $x_4=2^{a_4}{.5}^{b_4}{.7}^{c_4}{.11}^{d_4}$

Now, $x_1{x}_2{x}_3{x}_4=2^{a_1+a_2+a_3+a_4}{.5}^{b_1+b_2+b_3+b_4}{.7}^{c_1+c_2+c_3+c_4}{.11}^{d_1+d_2+d_3+d_4}$

As per given condition $a_1+a_2+a_3+a_4=1,$ which can be in $4$ ways $\left(1,0,0,0\right),\left(0,1,0,0\right)\left(0,0,1,0\right)$ and $\left(0,0,0,1\right),$ similarly for other powers.

Hence total number of ways $=4\times 4\times 4\times 4=256$