Question

If $^{n}C_{r-1}=28$, $^{n}C_{r}=56$ and $^{n}C_{r+1}=70$, then the value of $r$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The Correct Option is C

${ }^{n} C_{r-1}=28,{ }^{n} C_{r}=56 \cdot{ }^{n} C_{r+1}=70$
$\Rightarrow\, \frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{56}{28}$
$\Rightarrow \, \frac{\frac{n !}{r(r-1) !(n-r) !}}{\frac{n !}{(r-1) !(n-r+1)(n-r) !}}{=n-r+1=2 r}$
$\Rightarrow\, n-3 r=-1\,\,\,\,\,\dots(i)$
$\frac{{ }^{n} C_{r+1}}{{ }^{n} C_{r}}=\frac{70}{56}$
$\Rightarrow\, \frac{n !}{\frac{(r+1) !(n-r-1) !}{n !}}{\frac{n !}{r !(n-r) !}}=\frac{35}{28}$
$\Rightarrow\, \frac{\frac{(r+1) r !(n-r-1) !}{n !}}{\frac{35}{r !(n-r)(n-r-1) !}}=\frac{35}{28}$
$\Rightarrow\,\frac{n-r}{r+1}=\frac{35}{28}$
$\Rightarrow \, 28 n-28 r=35 r+35$
$\Rightarrow\, 28 n-63 r=35\,\,\,\,\,\,\dots(ii)$
Multiply by 21 in E (i) and subtracting from E (ii),
$21 n-63 r=-21$
$28 n-63 r=35$
$\frac{-\quad+\quad-}{-7 n=-56 \Rightarrow n=8}$
From E (i), $3 r=n+1=8+1$
$r=3$