Combinations

If the number of elements increase by 3, it increases the number of combinations of the second class of these elements 5 times. How many are the elements?

Correct result:

n =  3

Solution:

$C_{{ 2}}(6) = \dbinom{ 6}{ 2} = \dfrac{ 6! }{ 2!(6-2)!} = \dfrac{ 6 \cdot 5 } { 2 \cdot 1 } = 15 \ \\ 5 \cdot \ { { n } \choose 2 }={ { n+3 } \choose 2 } \ \\ 5 \cdot \ n \ \\ 5*n(n-1)=(n+3)*(n+2) \ \\ \ \\ 5 \cdot \ n(n-1)=(n+3) \cdot \ (n+2) \ \\ 4n^2 -10n -6=0 \ \\ \ \\ a=4; b=-10; c=-6 \ \\ D=b^2 - 4ac=10^2 - 4\cdot 4 \cdot (-6)=196 \ \\ D>0 \ \\ \ \\ n_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 10 \pm \sqrt{ 196 } }{ 8 } \ \\ n_{1,2}=\dfrac{ 10 \pm 14 }{ 8 } \ \\ n_{1,2}=1.25 \pm 1.75 \ \\ n_{1}=3 \ \\ n_{2}=-0.5 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 4 (n -3) (n +0.5)=0 \ \\ n>0 \ \\ n=n_{1}=3 \ \\ C_{1}={ { n } \choose 2 }=3 \ \\ C_{{ 2}}(3)=\dbinom{ 3}{ 2}=\dfrac{ 3! }{ 2!(3-2)!}=\dfrac{ 3 } { 1 }=3 \ \\ \ \\ C_{2}={ { n+3 } \choose 2 }={ { 3+3 } \choose 2 }=15 \ \\ n=3$

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