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The coefficients of $x^p$, and $x^q$ in the expansion of $(1+x)^{p+q}$ are

$\begin{array}{1 1} Equal \\ Equal\;with\;opposite\;sign \\ Reciprocal \;to\;each\;other \\ None\;of\;the\;above\;option \end{array}$

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General term in the expansion of $(1+x)^{p+q}$ is
$T_{r+1}=^{p+q}C_rx^r$
for coeff. of $x^p$, $r=p$ and for coeff. of $x^q$, $r=q$
$\therefore\:$coeff. of $x^p=^{p+q}C_p=\large\frac{(p+q)!}{q!.p!}$ and
coeff. of $x^q=^{p+q}C_q=\large\frac{(p+q)!}{q!.p!}$
$\therefore $ coeff. of $x^p$ = coeff. of $x^q$
answered Sep 26, 2013 by rvidyagovindarajan_1
 

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