Combinations

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

Correct answer:

n =  3

Step-by-step explanation:

$$\begin{gathered} C_2(6)=\left(\genfrac{}{}{0ex}{}{6}{2}\right)=\frac{6!}{2!(6-2)!}=\frac{6\cdot 5}{2\cdot 1}=15 \\ 5\cdot \left(\genfrac{}{}{0ex}{}{n}{2}\right)=\left(\genfrac{}{}{0ex}{}{n+3}{2}\right) \\ 5\cdot n\text{/}2 \\ 5\cdot n(n-1)=(n+3)\cdot (n+2) \\ 5\cdot n(n-1)=(n+3)\cdot (n+2) \\ 4n^2-10n-6=0 \\ 4=2^2 \\ 10=2\cdot 5 \\ 6=2\cdot 3 \\ \text{GCD}(4,10,6)=2 \\ 2n^2-5n-3=0 \\ a=2;b=-5;c=-3 \\ D=b^2-4ac=5^2-4\cdot 2\cdot (-3)=49 \\ D>0 \\ {n}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{5\pm \sqrt{49}}{4} \\ {n}_{1,2}=\frac{5\pm 7}{4} \\ {n}_{1,2}=1.25\pm 1.75 \\ {n}_1=3 \\ {n}_2=-0.5 \\ n>0 \\ n=n_1=3 \\ \text{ Verifying Solution: } \\ {C}_1=\left(\genfrac{}{}{0ex}{}{n}{2}\right)=3 \\ {C}_2(3)=\left(\genfrac{}{}{0ex}{}{3}{2}\right)=\frac{3!}{2!(3-2)!}=\frac{3}{1}=3 \\ {C}_2=\left(\genfrac{}{}{0ex}{}{n+3}{2}\right)=\left(\genfrac{}{}{0ex}{}{3+3}{2}\right)=15 \\ n=3 \end{gathered}$$

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