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. $4$
• E. $5$

Answer 1

Answer: (C)

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