You want to count the integer solutions of $$x_1+x_2+x_3+x_4+x_5=32$$ where $$x_i\gt3,\text{ i.e., }x_i\ge4.$$ If you substitute $$x_i=4+y_i$$ this becomes $$(4+y_1)+(4+y_2)+(4+y_3)+(4+y_4)+(4+y_5)=32\text{ with }4+y_i\ge4,$$ i.e., $$y_1+y_2+y_3+y_4+y_5=12\text{ with }y_i\ge0;$$ the number of solutions is of course$$\binom{16}4=1820.$$ If you prefer, you can substitute $$x_i=3+z_i$$ and the problem becomes $$(3+z_1)+(3+z_2)+(3+z_3)+(3+z_4)+(3+z_5)=32,\ 3+z_i\gt3,$$ i.e., $$z_1+z_2+z_3+z_4+z_5=17,\ z_i\gt0.$$