Question

If ${}^n{P}_{r+1}{:}^n{P}_{r+2}=11:52$, find the value of $n-r$

Answer 1

${}^n{P}_{r+1}{:}^n{P}_{r+2}=11:52$

$⟹\frac{n!}{(n-r-1)!}:\frac{n!}{(n-r-2)!}=11:52$

$⟹\frac{1}{(n-r-1)(n-r-2)!}:\frac{1}{(n-r-2)!}=11:52$

$⟹\frac{1}{n-r-1}=\frac{11}{52}$

$⟹n-r-1=\frac{52}{11}$

$⟹n-r=\frac{52}{11}+1$

$\therefore n-r=\frac{63}{11}$