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?

Final 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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