How many different circles is determined by 9 points at the plane, if 6 of them lie in a straight line?

Correct result:

n =  64


n1=C3(9)=(93)=9!3!(93)!=987321=84 n2=C3(6)=(63)=6!3!(63)!=654321=20 n=n1n2=8420=64n_1 = C_{{ 3}}(9) = \dbinom{ 9}{ 3} = \dfrac{ 9! }{ 3!(9-3)!} = \dfrac{ 9 \cdot 8 \cdot 7 } { 3 \cdot 2 \cdot 1 } = 84 \ \\ n_2 = C_{{ 3}}(6) = \dbinom{ 6}{ 3} = \dfrac{ 6! }{ 3!(6-3)!} = \dfrac{ 6 \cdot 5 \cdot 4 } { 3 \cdot 2 \cdot 1 } = 20 \ \\ n = n_1 - n_2 = 84 - 20 = 64

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