Points in space

There are n points, of which no three lie on one line and no four lies on one plane. How many planes can be guided by these points? How many planes are there if there are five times more than the given points?

Correct result:

r =  10
r₂ =  35

Solution:

$$\begin{gathered} C_3(10)=\left({\displaystyle \genfrac{}{}{0px}{}{10}{3}}\right)={\displaystyle \frac{10!}{3!(10-3)!}}={\displaystyle \frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}}=120 \\ n=10 \\ r=\left(\genfrac{}{}{0px}{}{n}{3}\right)=120=10 \end{gathered}$$

$$\begin{gathered} C_3(7)=\left({\displaystyle \genfrac{}{}{0px}{}{7}{3}}\right)={\displaystyle \frac{7!}{3!(7-3)!}}={\displaystyle \frac{7\cdot 6\cdot 5}{3\cdot 2\cdot 1}}=35 \\ 5n=\left(\genfrac{}{}{0px}{}{n}{3}\right) \\ 5n=n \\ 5n=n(n-1)(n-2)(n-3) \\ 5=(x-1)(x-2)\mathrm{/}6 \\ -0.166666666667x^2+0.5x+4.667=0 \\ 0.166666666667x^2-0.5x-4.667=0 \\ a=0.166666666667;b=-0.5;c=-4.667 \\ D=b^2-4ac=0.5^2-4\cdot 0.166666666667\cdot (-4.667)=3.3611111111 \\ D>0 \\ {x}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{0.5\pm \sqrt{3.36}}{0.333333333333}} \\ {x}_{1,2}=1.5\pm 5.5 \\ {x}_1=7 \\ {x}_2=-4 \\ \text{ Factored form of the equation: } \\ 0.166666666667(x-7)(x+4)=0 \\ {r}_2=\left(\genfrac{}{}{0px}{}{x_1}{3}\right)=35 \end{gathered}$$

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