AP and TS EAMCET Engineering Model Paper 18

Using prim e-factorization of 1050, we can write the given equation as
x1x2x3x4x5=2×3×52×7
we can assign 2,3 or 7 to any o f 5 variables. We can assign entire 52 to just one variable in 5 ways or can assign 52=5×5 to two variables in 5C2 ways.
Thus 52 can be assigned in 5C1+5C2=5+0=15‌ways
Thus, the required number of solutions is 5×5×5×15=1875