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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Find the coefficient of $a^5b^7$ in $(a-2b)^{12}$

$\begin{array}{1 1}(A)\;101376\\(B)\;-101376\\(C)\;102386\\(D)\;104386\end{array} $

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1 Answer

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
$(a-2b)^{12}=[a+(-2b)]^{12}$
General term $T_{r+1}=C(12,r)a^{12-r}(-2b)r$
Putting $12-r=5$
$12-5=r$
$\therefore r=7$
$T_{7+1}=C(12,7)a^{12-7}(-2b)^7$
$\Rightarrow C(12,7)a^5(-2)^7b^7$
$\Rightarrow C(12,7)(-2)^7a^5b^7$
Hence required coefficient is
$C(12,7)(-2)^7=-\large\frac{12!}{7!(12-7)!}$$.2^7$
$\Rightarrow \large\frac{-12\times 11\times 10\times 9\times 8\times 7!}{7!\times 5\times 4\times 3\times 2\times 1}$$\times 2^7$
$\Rightarrow -11\times 9\times 2^7$
$\Rightarrow -99\times 128$
$\Rightarrow -101376$
Hence (B) is the correct answer.
answered Jun 23, 2014 by sreemathi.v
 
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