\[n\]
is \[{}^nC_3 - {}^{2n}C_2 \gt 0\]
?\[{}^nC_3= \frac{n!}{3!(n-3)!}=\frac{n(n-1)(n-2)}{6}\]
\[{}^{2n}C_2= \frac{(2n)!}{2!(2n-2)!}=\frac{2n(2n-1)}{2}\]
Then
\[\frac{n(n-1)(n-2)}{6} - \frac{2n(2n-1)}{2} \gt 0\]
\[\frac{n}{6}((n-1)(n-2)-6(2n-1)) \gt 0\]
\[\frac{n}{6}(n^2-3n+2-12n+6) \gt 0\]
\[\frac{n}{6}(n^2-15n+8) \gt 0\]
The quadratic inside the brackets is positive for
\[n \ge 15\]
so (1) is true for \[n \ge 15\]
.