1.

If `r lt s le n " then prove that " ^(n)P_(s) " is divisible by "^(n)P_(r).`

Answer» Let s=r+k where `0 le k le s -r`. Then,
`.^(n)P_(s)=(n!)/((n-s)!)`
`=n(n-1)(n-2)..(n-(s-1))`
`=n(n-1)(n-2)..(n-(r+k-1))`
`=n(n-1)(n-2)..(n-(r-1))(n-r)(n-(r+1))..(n-(r+k-1))`
`={n(n-1)(n-2)..n-(r-1)}{(n-r)(n-(r+1))..(n-(r+k-1))}`
`= .^(n)P_(r ){(n-r)(n-(r+1))..(n-(r+k-1))}`
`= . ^(n)P_(r )xx " Integer"`
Hence, `.^(n)P_(s)` is divisible by `.^(n)P_(r )`


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