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

Answer»

Solution :`.^(N)P_(r) = (n!)/((n-r)!)`
`=n (n-1) (n-2)... (n-r+1)`
`.^(n)P_(s) = (n!)/((n-s)!)`
`= n(n-1)(n-2) ...(n-r+1)`
`(n-r) ...(n-s+1) ( :' r le s)`
Now `(.^(n)P_(s))/(.^(n)P_(r)) = (n-r) (n-r-1).......(n-s+1)`
= a positive INTEGER.
`:. .^(n)P_(s)` is DIVIDED by `.^(n)P_(r)` Hence PROVED.


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