# Combinations

How many elements can form six times more combinations fourth class than combination of the second class?

Correct result:

n =  11

#### Solution:

$C_{{ 2}}(11) = \dbinom{ 11}{ 2} = \dfrac{ 11! }{ 2!(11-2)!} = \dfrac{ 11 \cdot 10 } { 2 \cdot 1 } = 55 \ \\ C_{{ 4}}(11) = \dbinom{ 11}{ 4} = \dfrac{ 11! }{ 4!(11-4)!} = \dfrac{ 11 \cdot 10 \cdot 9 \cdot 8 } { 4 \cdot 3 \cdot 2 \cdot 1 } = 330 \ \\ 6 \cdot \ { { n } \choose 2 }={ { n } \choose 4 } \ \\ 6 \cdot \ n \ \\ 3 /((n-2)(n-3)(n-4)!)=1/(4 \cdot \ 3 \cdot \ 2)/(n-4)! \ \\ 4*3*2 * 3=(n-2)(n-3) \ \\ \ \\ 4 \cdot \ 3 \cdot \ 2 \cdot \ 3=(n-2)(n-3) \ \\ -n^2 +5n +66=0 \ \\ n^2 -5n -66=0 \ \\ \ \\ a=1; b=-5; c=-66 \ \\ D=b^2 - 4ac=5^2 - 4\cdot 1 \cdot (-66)=289 \ \\ D>0 \ \\ \ \\ n_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 5 \pm \sqrt{ 289 } }{ 2 } \ \\ n_{1,2}=\dfrac{ 5 \pm 17 }{ 2 } \ \\ n_{1,2}=2.5 \pm 8.5 \ \\ n_{1}=11 \ \\ n_{2}=-6 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (n -11) (n +6)=0 \ \\ n>0 \ \\ n=n_{1}=11 \ \\ C_{1}={ { n } \choose 2 }=55 \ \\ C_{2}={ { n } \choose 4 }=330$

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