Coefficient of $${t^{24}}$$ in $${\left( {1 + {t^2}} \right)^{12}}\left( {1 + {t^{12}}} \right)\left( {1 + {t^{24}}} \right)$$ is

The sum $$\sum\limits_{i = 0}^m {\left( {\matrix{ {10} \cr i \cr } } \right)\left( {\matrix{ {20} \cr {m - i} \cr } } \right),\,\left( {where\left( {\matrix{ p \cr q \cr } } \right) = 0\,\,if\,\,p < q} \right)} $$ is maximum when $$m$$ is

In the binomial expansion of $${\left( {a - b} \right)^n},\,n \ge 5,$$ the sum of the $${5^{th}}$$ and $${6^{th}}$$ terms is zero. Then $$a/b$$ equals

For $$2 \le r \le n,\,\,\,\,\left( {\matrix{ n \cr r \cr } } \right) + 2\left( {\matrix{ n \cr {r - 1} \cr } } \right) + \left( {\matrix{ n \cr {r - 2} \cr } } \right) = $$