solve eq nC3 +nC4=n(n-2)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use the following factorial formula, such that:

`C_n^k = (n!)/(k!(n-k)!)`

Reasoning by analogy, yields:

`C_n^3 = (n!)/(3!(n-3)!)`

`C_n^4 = (n!)/(4!(n-4)!)`

You need to re-write the equation using the factorial formula, such that:

`(n!)/(3!(n-3)!) + (n!)/(4!(n-4)!) = n(n -2)`

You need to use the following factorial relations between denominators such that:

`4! = 3!*4`

`(n - 3)! = (n - 4)!*(n - 3)`

`(n!)/(3! (n - 4)!*(n - 3)) + (n!)/(3!*4(n - 4)!) = n(n -2)`

Bringing the terms to a common denominator, yields:

`(n!*4 + n!*(n - 3))/(3! (n - 4)!*(n - 3)) + (n!)/(3!*4(n - 4)!) = n(n -2)`

Factoring out `n!` to numerator, such that:

`(n!*(4 + n - 3))/(4!*(n - 3)!) = n(n - 2)`

`(n!*(n + 1))/(4!*(n - 3)!) = n(n - 2)`

Using the following factorial relation between numerator and denominator, yields:

`((n - 3)!(n - 2)*(n - 1)*n*(n + 1))/(4!*(n - 3)!) = n(n - 2)`

Reducing duplicate factors yields:

`((n - 2)*(n - 1)*n*(n + 1)) = 4!*n(n - 2)`

Moving all terms to one side yields:

`((n - 2)*(n - 1)*n*(n + 1)) - 1*2*3*4*n*(n -2) = 0`

Factoring out `n(n - 2)` yields:

`n(n - 2)(n^2 - 1 - 24) = 0`

Using zero product rule, yields:

`{(n(n - 2) = 0),(n^2 -25 = 0):} => {(n = 0, n = 2),(n = +-5):}`

Since n needs to be a natural number and it needs to be larger or equal to 4, yields that the value accepted is `n = 5` .

Hence, evaluating the solution to the given equation, yields `n = 5.`

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