Combinations 80637

If the number of elements decreases by 4, the number of combinations of the second class from these elements decreases three times. How many elements are there?

Correct answer:

n =  10

Step-by-step explanation:

$$\begin{gathered} C_2(6)=\left(\genfrac{}{}{0ex}{}{6}{2}\right)=\frac{6!}{2!(6-2)!}=\frac{6\cdot 5}{2\cdot 1}=15 \\ \left(\genfrac{}{}{0ex}{}{n}{2}\right)/3=\left(\genfrac{}{}{0ex}{}{n-4}{2}\right) \\ \frac{n!}{2\cdot (n-2)!}=3\cdot \frac{(n-4)!}{2\cdot (n-6)!} \\ \frac{n(n-1)(n-2)!}{(n-2)!}=3\cdot \frac{(n-4)(n-5)(n-6)!}{(n-6)!} \\ n(n-1)=3\cdot (n-4)\cdot (n-5) \\ n(n-1)=3\cdot (n-4)\cdot (n-5) \\ -2n^2+26n-60=0 \\ 2n^2-26n+60=0 \\ 2...\text{ prime number} \\ 26=2\cdot 13 \\ 60=2^2\cdot 3\cdot 5 \\ \text{GCD}(2,26,60)=2 \\ {n}^2-13n+30=0 \\ a=1;b=-13;c=30 \\ D=b^2-4ac=13^2-4\cdot 1\cdot 30=49 \\ D>0 \\ {n}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{13\pm \sqrt{49}}{2} \\ {n}_{1,2}=\frac{13\pm 7}{2} \\ {n}_{1,2}=6.5\pm 3.5 \\ {n}_1=10 \\ {n}_2=3 \\ n>3 \\ n=n_1=10 \\ \text{ Verifying Solution: } \\ {c}_1=\left(\genfrac{}{}{0ex}{}{n}{2}\right)=45 \\ {C}_2(10)=\left(\genfrac{}{}{0ex}{}{10}{2}\right)=\frac{10!}{2!(10-2)!}=\frac{10\cdot 9}{2\cdot 1}=45 \\ {c}_2=\left(\genfrac{}{}{0ex}{}{n-4}{2}\right)=\left(\genfrac{}{}{0ex}{}{10-4}{2}\right)=15 \\ 3\cdot c_2=c_1 \end{gathered}$$

Our quadratic equation calculator calculates it.

Try another example