What are natural n>=2 in equation nC1+nC2=120?

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You need to replace the binomial coefficients by factorial forms, such that:

`nC1 = (n!)/(1!(n - 1)!) => nC1 = ((n-1)!*n)/(1!(n - 1)!)`

Reducing duplicate factors yields:

`nC1 = n`

`nC2 = (n!)/(2!(n - 2)!) => nC2 = ((n-2)!*(n-1)*n)/(2(n - 2)!)`

Reducing duplicate factors yields:

`nC2 = ((n-1)*n)/2`

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

`n + ((n-1)*n)/2 = 120`

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

`2n + n^2 - n - 240 = 0`

`n^2 + n - 240 = 0`

You should use quadratic formula, such that:

`n_(1,2) = (-1+-sqrt(1+960))/2 => n_(1,2) = (-1+-31)/2`

`n_1 = 15; n_2 = -16`

Since the problem indicates to accept the natural solutions, hence, only `n = 15` is valid.

**Hence, evaluating the natural solutions to factorial equation yields `n = 15` .**

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