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If ${ }^{n} C_{r-1}=36,{ }^{n} C_{r}=84$ and ${ }^{n} C_{r+1}=126$ then the value of ${ }^{n} C_{8}$ is
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Verified Answer
The correct answer is:
9
Given,
$$
\begin{aligned}
{ }^{n} C_{r-1} &=36 \\
{ }^{n} C_{r} &=84 \\
{ }^{n} C_{r+1} &=126
\end{aligned}
$$
On dividing
$$
\frac{r}{n-r+1}=\frac{36}{84}
$$
$\Rightarrow \quad 84 r=36 n-36 r+36$
$\Rightarrow \quad 120 r=36 n+36$
Also,
$$
\begin{array}{r}
\frac{r+1}{n-r}=\frac{64}{126} \\
\Rightarrow \quad 126 r+126=84 n-84 r
\end{array}
$$
$\Rightarrow$
$$
210r=84 n-126
$$
On solving Eqs. (iv) and (v). we get $n=9$ and $r=3$
\begin{array}{l}
\hline
\end{array}
$$
{ }^{n} C_{8}={ }^{9} C_{8}=9
$$
$$
\begin{aligned}
{ }^{n} C_{r-1} &=36 \\
{ }^{n} C_{r} &=84 \\
{ }^{n} C_{r+1} &=126
\end{aligned}
$$
On dividing
$$
\frac{r}{n-r+1}=\frac{36}{84}
$$
$\Rightarrow \quad 84 r=36 n-36 r+36$
$\Rightarrow \quad 120 r=36 n+36$
Also,
$$
\begin{array}{r}
\frac{r+1}{n-r}=\frac{64}{126} \\
\Rightarrow \quad 126 r+126=84 n-84 r
\end{array}
$$
$\Rightarrow$
$$
210r=84 n-126
$$
On solving Eqs. (iv) and (v). we get $n=9$ and $r=3$
\begin{array}{l}
\hline
\end{array}
$$
{ }^{n} C_{8}={ }^{9} C_{8}=9
$$
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