Find r if `.^(5)P_(r )= .^(6)P_(r-1)`

Explanation

`(5!)/((5-r)!)=(6!)/([6-(r-1)]!)rArr (5!)/((5-r)!)=(6xx5!)/((7-r)!)`
`therefore " " (7-r)!=6(5-r)!`
`(7-r)(6-r)(5-r)!=6 (5-r)!`
`therefore " " (7-r) (6-r) = 6`
`therefore " " r^(2)-13r+36 = 0 rArr r = 4, 9`
r = 9 is not possible `" " rArr r = 4`