# Combinations

From how many elements we can create 990 combinations 2nd class without repeating?

Correct result:

n =  45

#### Solution:

${{ n} \choose 2} = \dfrac{n(n-1)}{2}=990;n>0 \ \\ n^2 -n -1980 =0 \ \\ \ \\ a=1; b=-1; c=-1980 \ \\ D = b^2 - 4ac = 1^2 - 4\cdot 1 \cdot (-1980) = 7921 \ \\ D>0 \ \\ \ \\ n_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 1 \pm \sqrt{ 7921 } }{ 2 } \ \\ n_{1,2} = \dfrac{ 1 \pm 89 }{ 2 } \ \\ n_{1,2} = 0.5 \pm 44.5 \ \\ n_{1} = 45 \ \\ n_{2} = -44 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (n -45) (n +44) = 0 \ \\ \ \\ \ \\ n>0 \Rightarrow n=45$

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