We first expand each of the factors of the given product using binomial theorem. We have $(1+2x)^6$ = $\binom{6}{0} + \binom{6}{1}(2x) + \binom{6}{2}(2x)^2$ + $\binom{6}{3}(2x)^3 + \binom{6}{4}(2x)^4 + \binom{6}{5}(2x)^5$ + $\binom{6}{6}(2x)^6$ = 1 + 12x + $60x^2+160x^3$ + $240x^3+240x^4+192x^5+64x^6$ and $(1-x)^7$ = $\binom{7}{0} + \binom{7}{1}(-x) + \binom{7}{2}(-x)^2$ + $\binom{7}{3}(-x)^3 + \binom{7}{4}(-x)^4$ + $\binom{7}{5}(-x)^5 + \binom{7}{6}(-x)^6 + \binom{7}{7}(-x)^7$ = 1 - 7x + $21x^2-35x^3+35x^4-21x^5$ + $7x^6-x^7$ Thus $(1+2x)^6 (1-x)^7$ = (1 + 12x + $60x^2+160x^3+240x^3$ + $240x^4+192x^5+64x^6$) × (1 - 7x + $21x^2-35x^3+35x^4-21x^5$ + $7x^6-x^7$) We write only those terms which involves $x^5$. This can be done if we note, that $x^r \cdot x^{5-r}$ = $x^5$. The terms containing $x^5$ are 1 $(-21x^5) + 12x(35x^4)$ + $60x^2 (-35x^3)+160x^3(21x^2)$ + $240x^4 (-7x) +192x^5 (1)$ = $-21x^5 + 420 x^5 - 2100 x^5$ + $3360x^5 - 1680x^5 + 192x^5$ = $171x^5$ Thus the coefficients of x5 in the given product is 171.