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If .^(n)C(r)=.^(n)C(r-1) and .^(n)P(r)=....

If `.^(n)C_(r)=.^(n)C_(r-1) and .^(n)P_(r)=.^(r)P_(r+1)`, the value of n is

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`.^(n)C_(r) = .^(n)C_(r-1)`
`rArr n = r +(r-1)`
`rArr n = 2r - 1` ...(1)
and `.^(n)P_(r)- .^(n)P_(r+1)`
`rArr (n!)/((n-r)!) =(n!)/((n-r-1)!)`
`rArr (n-r)! = (n-r-1)!`
`rArr (n-r)(n-r-1)! = (n-r-1)!`
`rArr n- r = 1`
`rARr 2r - 1 - r = 1` [From eq. (1)]
`rArr r = 2`
`rArr` From eq. (1) `n = 2r - 1`
`=2(2) - 1 = 3`
`n = 3, r = 2`
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