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