Consider the following statements

I) The number of non trivial even divisors of the number $2^{{\alpha }_1} 3^{{\alpha }_2} 4^{{\alpha }_3} 5^{{\alpha }_4} 6^{{\alpha }_5}$ is $\left({\alpha }_1+1\right)\left({\alpha }_2+1\right)\left({\alpha }_3+1\right)\left({\alpha }_4+1\right){\left({\alpha }_5+1\right)}^{-2}$

II) The number of non trivial odd divisiors of the number $2^{{\alpha }_1} 3^{{\alpha }_2} 4^{{\alpha }_3} 5^{{\alpha }_4} 6^{{\alpha }_5}$ is $({\mathrm{\alpha }}_2+{\mathrm{\alpha }}_5+1)({\mathrm{\alpha }}_4+1)-1$

$\mathrm{I}: \mathrm{N}=2^{\mathrm{\alpha }1} 3^{\mathrm{\alpha }2} 4^{\mathrm{\alpha }3} 5^{\mathrm{\alpha }4} 6^{\mathrm{\alpha }5}$

$\mathrm{N}=2^{\mathrm{\alpha }1} 3^{\mathrm{\alpha }2} 2^{2\mathrm{\alpha }3} 5^{\mathrm{\alpha }4} 2^{\mathrm{\alpha }5} 3^{\mathrm{\alpha }5}$

$=2^{{\mathrm{\alpha }}_1+2{\mathrm{\alpha }}_3+{\mathrm{\alpha }}_5} 3^{{\mathrm{\alpha }}_2+{\mathrm{\alpha }}_5} 5^{{\mathrm{\alpha }}_4}$

$\mathrm{No}. \mathrm{of} \mathrm{non} \mathrm{trivial} \mathrm{even} \mathrm{divisors}=({\mathrm{\alpha }}_1+2{\mathrm{\alpha }}_3+{\mathrm{\alpha }}_5)({\mathrm{\alpha }}_2+{\mathrm{\alpha }}_5+1)({\mathrm{\alpha }}_4+1)$

$\mathrm{II}: \mathrm{N}=2^{\mathrm{\alpha }1} 3^{\mathrm{\alpha }2} 4^{\mathrm{\alpha }3} 5^{\mathrm{\alpha }4} 6^{{\alpha }_5}$

$\mathrm{N}=2^{{\mathrm{\alpha }}_1+2{\mathrm{\alpha }}_3+{\mathrm{\alpha }}_5} 3^{{\mathrm{\alpha }}_2+{\mathrm{\alpha }}_5} 5^{{\mathrm{\alpha }}_4}$

$\mathrm{No}.\mathrm{of} \mathrm{odd} \mathrm{divisors}=({\mathrm{\alpha }}_2+{\mathrm{\alpha }}_5+1)({\mathrm{\alpha }}_4+1)$

$\mathrm{No}.\mathrm{of} \mathrm{odd} \mathrm{non} \mathrm{trivial} \mathrm{divisors} = ({\mathrm{\alpha }}_2+{\mathrm{\alpha }}_5+1)({\mathrm{\alpha }}_4+1)-1$