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

Explanation

We have
` 5 xx ""^(4)P_(r) =6 xx ""^(5)P_(r-1) `
` rArr 5 xx (4!)/((4-r)!) =6 xx (5!)/({5-(r-1)}!) `
` rArr (5!)/((4-r)!)=(6xx (5!))/((6-r)!) " " [ because 5 xx (4!)=5!] `
` rArr (1)/((4-r)!)=(6)/((6-r)!) rArr (1)/((4-r)!)=(6)/((6-r)(5-r){(4-r)!}) `
` rArr (6)/((6-r)(5-r))=1 rArr (6-r)(5-r)-6=0 `
` rArr r^(2)-11r+24=0 rArr (r-3)(r-8)=0 `
` rArr r=3 " " [ because r ne 8, "as " ""^(4)P_(8) " is not defined "]. `
Hence, ` r=3. `