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Solve equation xC3=5/4x(x-3)? combination C

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drifterkay | Student, Undergraduate | eNoter

Posted September 1, 2012 at 1:26 PM via web

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Solve equation xC3=5/4x(x-3)?

combination C

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted September 1, 2012 at 2:30 PM (Answer #1)

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`^nC_r = (n!)/[(r!)(n-r)!]`

 

`^xC_3 = (x!)/[3!(x-3)!]`

 

` (x!)/[3!(x-3)!] = (5/4)x(x-3)`

 

 

`x! = 1*2*3*......(x-3)(x-2)(x-1)x = (x-3)(x-2)(x-1)x`
 
 
`(x!)/[3!(x-3)!]`
`= (x-3)!(x-2)(x-1)x/[3!(x-3)!]`
`= (x-2)(x-1)x/(3!)`
 
 
`(x-2)(x-1)x/(3!)`     = `(5/4)x(x-3)`
    `(x-2)(x-1)x` = `1*2*3*(5/4)x(x-3)`
  
     `(x-2)(x-1)` = `15/2(x-3)`
 ` 2(x^2-3x+2)`   = `15x-45`
`2x^2-21x+49`  = 0
 
 
`x = (-(-21)+-sqrt((-21)^2-4*2*49))/(2*2)`
 
Solving this yields;
x = 7 and x = 3.5
 
But x is a part of combination. So `x in Z^+`
 
Therefore the answer is x = 7

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