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The coefficient if $x^{10}$ in the expansion $(1+x^2-x^3)^8$ is ?

\[\begin {array} {1 1} (a)\; 476 \\ (b)\; 496 \\ \ (c) \;506 \\ \ (d) \;528 \end {array}\]
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1 Answer

$(1+x^2-x^3)^8=\big[1+x^2(1-x)\big]^8$
$=^8C_0+^8C_1.x^2(1-x)+^8C_2.x^4(1-x)^2+$
$^8C_3.x^6(1-x)^3+^8C_4.x^8(1-x)^4+^8C_5x^{10}(1-x)^5+......$
It is clear that the terms containing $x^{10}$ are $^8C_4.x^8(1-x)^4$
and $^8C_5.x^{10}(1-x)^4$
$\therefore$ Coeff. of $x^{10}$ = $ ^8C_4$.[coeff. of $x^2$ in $(1-x)^4]+^8C_5$
.
$=^8C_4.6+^8C_5=70\times 6+56=476$
answered Oct 21, 2013 by rvidyagovindarajan_1
 

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