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Solve for n if C(n+2,4)=6C(n,2)
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You need to remember the factorial formula for combinations such that:
C_n^k = (n!)/(k!*(n-k)!)
Hence, C_(n+2)^4 = ((n+2)!)/(4!*(n+2-4)!)
C_(n+2)^4 = ((n+2)!)/(4!*(n-2)!)
C_n^2 = (n!)/(2!*(n-2)!)
Hence, writing the equation in factorial form yields:
((n+2)!)/(4!*(n-2)!) = 6(n!)/(2!*(n-2)!)
Reducing by (2!*(n-2)!) yields:
((n+2)!)/(3*4) = 6(n!) => (n+2)! = 12*6*n!
You may write (n+2)! in terms of n! such that:
(n+2)! = n!*(n+1)(n+2)
Writing the equation using (n+2)! = n!*(n+1)(n+2) yields:
`n!*(n+1)(n+2) = 12*6*n!`
`` Reducing by n! yields:
`(n+1)(n+2) = 72`
You need to open the brackets to the left side such that:
`n^2 + 2n + n + 2 = 72`
Subtracting 72 both sides yields:
`n^2 + 3n - 70 = 0`
You need to remember quadratic formula that helps finding the zeroes of quadratic equations.
`n_(1,2) = (-3+-sqrt(9 + 280))/2 =gt n_(1,2) = (-3+-sqrt(289))/2`
`n_(1,2) = (-3+-17)/2 =gt n_1 = (-3+17)/2 = 7`
`n_2 = (-3-17)/2 = -10`
Since n need to be positive, hence the only solution to the equation is n = 7.
Posted by sciencesolve on January 14, 2012 at 1:24 AM (Answer #1)
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