Find the coefficient of $x^{5}$ in the product $(1+2 x)^{6}(1-x)^{7}$ using binomial theorem.
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 $171 .$
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