If the value of  $55C_5-5C_{1}44C_5+5C_{2}33C_5-5C_{3}22C_5+5C_{4}11C_5$ is  $\left(a^b\right)$   where a and b both are  prime numbers, then the value of (a-b) = 

Given expression is coeff of  $x^5$ in
 ${(1+x)}^{55}-5C_1{(1+x)}^{44}+5C_2{(1+x)}^{33}-5C_3{(1+x)}^{22}+5C_4{(1+x)}^{11}-5C_5$
Put  ${(1+x)}^{11}=t$
 $t^5-5C_1{t}^4+5C_2{t}^3-5C_3{t}^2+5C_{4}t-5C_5$
 $\begin{array}{l}{(t-1)}^5\\ {[{(1+x)}^{11}-1]}^5\end{array}$
 $\begin{array}{l}{[\cancel{1}+11C_{1}x+11C_2{x}^2+\mathrm{.......}+11C_{11}{x}^{11}-\cancel{1}]}^5\\ x^5{[11C_1+11C_{2}x+\mathrm{.....}+11C_{11}{x}^{10}]}^5\end{array}$
Coeff of   $x^5$ is  ${11}^5$