Number of functions defined from $f:\{1,2,3,4,5,6\}\to \{7,8,9,10\}$ such that the sum $f\left(1\right)+f\left(2\right)+f\left(3\right)+f\left(4\right)+f\left(5\right)+f\left(6\right)$ is odd is

Giving any random configuration for f(1) to f(5) number of functions is $4^5$

Now the sum $\sum _{r=1}^5 f\left(r\right)$ is either odd (or) even. If it is odd then we have only two choices for f(6) i.e. the elements 8 to 10 if the sum is even then also we have only two choices for f(6) hence the total functions $=2^{11}\text{. }$