# What is sum of roots of equation nC3+nC4 = n(n-2)?

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### 1 Answer

You should use the following factorial formulas to express the terms `C_n^3` and `C_n^4` such that:

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

You may write n! in terms of (n - 3)!, such that:

`n! = (n - 3)!(n - 2)(n - 1)n`

`C_n^3 = ((n - 3)!(n - 2)(n - 1)n)/(3!(n - 3)!)`

Reducing duplicate factors yields:

`C_n^3 = ((n - 2)(n - 1)n)/(1*2*3)`

`C_n^3 = ((n - 2)(n - 1)n)/6`

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

`C_n^4 = ((n - 4)!(n - 3)(n - 2)(n - 1)n)/(4!(n - 4)!) `

Reducing duplicate factors yields:

`C_n^4 = ((n - 3)(n - 2)(n - 1)n)/(1*2*3*4) `

You need to re-write the equation in its simplified form, such that:

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

You need to bring all terms to a common denominator, such that:

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

`4n(n - 1)(n - 2) + n(n - 1)(n - 2)(n - 3) - 24n(n - 2) = 0`

You need to factor out `n(n - 2),` such that:

`n(n - 2)(4n - 4 + (n - 1)(n - 3) - 24) = 0`

`n(n - 2)(4n - 4 + n^2 - 4n + 3 - 24) = 0`

Reducing like terms yields:

`n(n - 2)(n^2 - 25) = 0`

Using zero product rule, yields:

`n = 0`

`n - 2 = 0 => n = 2`

`n^2 - 25 = 0 => n^2 = 25 => n_(1,2) = +-5`

You cannot accept all values for n since they need to be natural and they need to be larger than 3, but not equal to `3` , since the term `C_3^4` would be invalid.

**Hence, because of these enunciated conditions, you need to select the values that are valid and there exists only one value, `n = 5` thus, the sum of solutions to the given equation coincides to value of the root, `s = 5` .**