$4 \{^nC_1 + 4 . ^nC_2 + 4^2 . ^nC_3 + ...... + 4^{n - 1}\}$ ની કિમત મેળવો 

$x^2(1+x)^{98}+x^3(1+x)^{97}+x^4(1+x)^{96}+\ldots+x^{54}(1+x)^{46}$ ના વિસ્તરણમાં $x^{70}$ નો સહગુણક ${ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}}$ છે. તો $p+q$ ની શક્ય કિંમત ........... છે. 

$\left( {\left( {\begin{array}{*{20}{c}}
{21}\\
1
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
1
\end{array}} \right)} \right) + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
2
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
2
\end{array}} \right)} \right)$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
3
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
3
\end{array}} \right)} \right) + \;.\;.\;.$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
{10}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
{10}
\end{array}} \right)} \right) = $

જો ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .......... + {C_n}{x^2},$ તો $C_0^2 + C_1^2 + C_2^2 + C_3^2 + ...... + C_n^2$ =

જો $\sum\limits_{K = 1}^{12} {12K{.^{12}}{C_K}{.^{11}}{C_{K - 1}}} $ ની કિમત $\frac{{12 \times 21 \times 19 \times 17 \times ........ \times 3}}{{11!}} \times {2^{12}} \times p$ હોય તો $p$ ની કિમત મેળવો 

જો ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .... + {C_n}{x^n}$, તો ${C_0} + 2{C_1} + 3{C_2} + .... + (n + 1){C_n}$ = . . .