Question

$\left(i\right)$ If ${}^{m+r}{P}_2=90$ and ${}^{m-r}{P}_2=30$.
$\left(ii\right)$ If ${}^n{P}_r{=}^n{P}_{r+1}$ and ${}^n{C}_r={}^n{C}_{r-1}$.
Find $m+n-r$ ?

Answer 1

$\left(i\right)$ Since ${}^{m+r}{P}_2=90$
$\therefore (m+r)(m+r-1)=90$
$\iff (m+r{)}^2-(m+r)-90=0$
$\therefore m+r=10$ and $m+r=-9$
Hence $m+r=10,m+r\ne -9....\left(1\right)$
${}^{m-r}{P}_2=30$
$\therefore (m-r)(m-r-1)=30$
$\therefore (m-r{)}^2-(m-r)-30=0$
$\iff m-r=6,m-r=-5....\left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$, we get $m=8$ & $r=2$
Hence, $(m,r)=(8,2)$
$\left(ii\right)$ Given ${}^n{P}_r{=}^n{P}_{r+1}$
$\iff \frac{n!}{(n-r)!}=\frac{n!}{(n-r-1)!}$
$\iff (n-r)!=(n-r-1)!$
$\iff (n-r)(n-r-1)!-(n-r-1)!=0$
$\iff (n-r-1)!(n-r-1)=0$
$(n-r-1)!\ne 0\therefore n-r-1=0$
$n-r=1...\left(1\right)$
and given ${}^n{C}_r={}^n{C}_{r-1}\{\because {}^n{C}_A={}^n{C}_B\Rightarrow n=A+BorA=B\}$
$n=r+(r-1)$
$n-2r=-1...\left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$, we get
$r=2$ and $n=3$
$\therefore m+n-r=9$