Combinations - elements

If we increase the number of elements by 1, the number of combinations of the third class without repetitions increases by 10. How many elements do we have?

Correct answer:

n =  5

Step-by-step explanation:

$$\begin{gathered} C_3(6)=\left(\genfrac{}{}{0ex}{}{6}{3}\right)=\frac{6!}{3!(6-3)!}=\frac{6\cdot 5\cdot 4}{3\cdot 2\cdot 1}=20 \\ \left(\genfrac{}{}{0ex}{}{n+1}{3}\right)=10+\left(\genfrac{}{}{0ex}{}{n}{3}\right) \\ \frac{(n+1)!}{3!\cdot (n+1-3)!}=10+\frac{n!}{3!\cdot (n-3)!} \\ \frac{(n+1)!}{6\cdot (n-2)!}=10+\frac{n!}{6\cdot (n-3)!} \\ \frac{(n+1)n(n-1)(n-2)!}{6\cdot (n-2)!}=10+\frac{n(n-1)(n-2)(n-3)!}{6\cdot (n-3)!} \\ \frac{(n+1)n(n-1)}{6}=10+\frac{n(n-1)(n-2)}{6} \\ (n+1)n(n-1)=60+n(n-1)(n-2) \\ 3n^2-3n-60=0 \\ {n}_1=-4 \\ {n}_2=5 \\ n=5 \\ \text{ Verifying Solution: } \\ {C}_3(5)=\left(\genfrac{}{}{0ex}{}{5}{3}\right)=\frac{5!}{3!(5-3)!}=\frac{5\cdot 4}{2\cdot 1}=10 \\ {c}_1=\left(\genfrac{}{}{0ex}{}{n+1}{3}\right)=\left(\genfrac{}{}{0ex}{}{5+1}{3}\right)=20 \\ {c}_2=\left(\genfrac{}{}{0ex}{}{n}{3}\right)=10 \end{gathered}$$

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