Using Binomial Theorem, the expressions, $(1+2 x)^{6}$ and $(1-x)^{7},$ can be expanded as

${(1 + 2x)^6} = {\,^0}{C_0} + {\,^0}{C_1}(2x) + {\,^0}{C_2}{(2x)^2} + {\,^0}{C_3}{(2x)^3} + {\,^0}{C_4}{(2x)^4} + {\,^0}{C_5}{(2x)^5} + {\,^0}{C_6}{(2x)^6}$

$=1+6(2 x)+15(2 x)^{2}+20(2 x)^{3}+15(2 x)^{4}+6(2 x)^{5}+(2 x)^{6}$

$=1+12 x+60 x^{2}+160 x^{3}+240 x^{4}+192 x^{5}+64 x^{6}$

${(1 – x)^7} = {\,^7}{C_0} – {\,^7}{C_1}(x) + {\,^7}{C_2}{(x)^2} – {\,^7}{C_3}{(x)^3}$    

               $ + {\,^7}{C_4}{(x)^4} – {\,^7}{C_5}{(x)^5} + {\,^7}{C_6}{(x)^6} – {\,^7}{C_7}{(x)^7}$

$=1-7 x+21 x^{2}-35 x^{3}+35 x^{4}-21 x^{5}+7 x^{6}-x^{7}$

$\therefore(1+2 x)^{6}(1-x)^{7}$

$=\left(1+12 x+60 x^{2}+160 x^{3}+240 x^{4}+192 x^{5}+64 x^{6}\right)$

              $\left(1-7 x+21 x^{2}-35 x^{3}+35 x^{4}-21 x^{5}+7 x^{6}-x^{7}\right)$

The complete multiplication of the two brackets is not required to be carried out. Only those terms, which involve $x^{5},$ are required.

The terms containing $x^{5}$ are

$1\left(-21 x^{5}\right)+(12 x)\left(35 x^{4}\right)+\left(60 x^{2}\right)\left(-35 x^{3}\right)+\left(160 x^{3}\right)\left(21 x^{2}\right)$

                $+\left(240 x^{4}\right)(-7 x)+\left(192 x^{5}\right)(1)$

$=171 x^{5}$

Thus, the coefficient of $x^{5}$ in the given product is $171 .$