Prove that:
(i) ` ""^(n)P_(n)=""^(n)P_(n-1) " (ii) "^(n)P_(r)=n* ""^(n-1)P_(r-1) " (iii) "^(n-1)P_(r)+r* ""^(n-1)P_(r-1)=""^(n)P_(r)`

Explanation

By using the formula for ` ""^(n)P_(r),` we have
(i) ` ""^(n)P_(n-1)=(n!)/({n-(n-1)]!)=(n!)/(1!)=n! =""^(n)P_(n).`
(ii) ` ""^(n)P_(r)=(n!)/((n-r)!)=(n*(n-1)!)/([(n-1)-(r-1)]!)=n* ""^(n-1)P_(r-1).`
(iii) ` ""^(n-1)P_(r)+r* ""^(n-1)P_(r-1)`
`={((n-1)!)/((n-1-r)!)+r*((n-1)!)/([(n-1)-(r-1)]!)}`
`={((n-1)!)/((n-1-r)!)+r*((n-1)!)/((n-r)!)}`
`={((n-1)!)/((n-r-1)!)+r*((n-1)!)/((n-r)*[(n-r-1)!])}`
`=((n-1)!)/((n-r-1)!){1+(r)/((n-r))}=(n*[(n-1)!])/((n-r)*[(n-r-1)!])`
`= (n!)/((n-r)!)= ""^(n)P_(r).`
Hence, `""^(n-1)P_(r)+r* ""^(n-1)P_(r-1)= ""^(n)P_(r).`