Find r if `.^(5)P_(r )=2 .^(6)P_(r-1)`

Explanation

`.^(5)P_(r )=2 .^(6)P_(r -1)rArr (5!)/((5-r)!)=2(6!)/((7-r)!)`
`(5!)/((5-r)!)=((2)(6)5!)/((7-r)(6-r)(5-r)!)`
`1=(12)/((7-r)(6-r))`
`(7-r)(6-r)=12`
`r^(2)-13r + 30 =0`
`(r-10)(r-3)=0`
`therefore r = 10, 3`.
but r = 10 is not possible because then `.^(5)P_(r )` is undefined `therefore r = 3`