# If β = nC2, then what is the value of βC2?

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You need to remember the factorial formula of combinations to calculate `beta` such that:

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

Reducing by `(n-2)!` yields:

`beta = (n(n-1))/2`

Hence, `betaC2 = (beta!)/(2!*(beta - 2)!) =gt betaC2 = ((beta-2)!*(beta-1)*beta)/(1*2*(beta-2)!) `

Reducing by `(beta-2)! ` yields: `betaC2 = beta*(beta - 1)/2`

Substituting `(n(n-1))/2` for `beta` yields:

`betaC2 = (n(n-1))(n(n-1)-2)/8 =gt betaC2 = (n^2(n-1)^2 - 2n^2 + 2n)/8`

`` `betaC2 = (n^4 - 2n^3 - n^2 + 2n)/8`

**Hence, evaluating `betaC2` under the condition `beta=nC2` yields `betaC2 = (n^4 - 2n^3 - n^2 + 2n)/8` .**