A probability problem by Khaled Barie

By expanding the combination in the first equation:

$\large \frac{^{n+1}P_{r+1}}{^{n+1}C_{r+1}} = 5040$

Therefore, $\large ^{n+1}P_{r+1} \times \frac{(r+1)!}{^{n+1}P_{r+1}} = 5040$

$\large (r+1)! = 7!$

$\large r+1 = 7$

$\large r = 6$

By expanding both combinations in the second equation:

$\Large \frac{(n+1)!}{(n+1-r)! \times r!} \times \frac{(n+1-r+1)! \times (r-1)!}{(n+1)!} = \frac{2}{3}$

$\Large \frac{(n+1-r+1)! \times (r-1)!}{(n+1-r)! \times r!}= \frac{2}{3}$

$\Large \frac{(n+2-r)! \times (r-1)!}{(n+1-r)! \times r!}= \frac{2}{3}$

$\Large \frac{(n+2-r) \times(n+1-r)! \times (r-1)!}{(n+1-r)! \times r \times (r-1)!}= \frac{2}{3}$

$\Large \frac{n+2-r}{r }= \frac{2}{3}$

Since $r=6$

$\Large \frac{n-4}{6}= \frac{2}{3}$

$\large 3n-12= 12$

$\large 3n= 24$

$\large n=8$

$\large n+r=8+6= \boxed{14}$