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

combination C

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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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