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If ^n Pr=^n P(r+1)a n d^n Cr=^n C(r-1,) ...

If `^n P_r=^n P_(r+1)a n d^n C_r=^n C_(r-1,)` then the value of `n+r` is.

Text Solution

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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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