Suppose $a^5b^7$ occurs in the (r + 1)th term of the expansion $(a-2b)^{12}$ We know that the (r + 1)th term in the expansion of $(x + a)^n$ is given by $T_{r+1}$ = $\\,{}^n C_r x^{n-r} a^r$ ⇒ $T_{r+1}$ = $\\,{}^{12} C_r (a)^{12-r}(-2b)^r$ = $\\,{}^{12} C_r a^{12-r} b^r (-2)^r$ ... (i) On comparing power of a as well as b in $a^5b^7$ and $T_{r+1}$ , we get 12 – r = 5 ⇒ 12 – 5 = r ⇒ r = 7 Putting r = 7 in (i), we obtain the coefficient of $a^5 b^7$ in the expansion of $(a-2b)^{12}$ = $\\,{}^{12} C_7 (-2)^7$ = $\frac{12!}{7!5!}$ × (- 128) = $\frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1}$ × (- 128) = - 101376