A vessel contains two non-reactive gases neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is $3:2$. the ratio of number of molecules is

As partial pressure of a gas in a mixture is the pressure it would exert for the same volume and temperature, if it alone occupied the vessel, therefore, for common $\mathrm{V}$ and $\mathrm{T}$, ${\mathrm{P}}_1$ and ${\mathrm{P}}_2$ are partial pressures

${\mathrm{P}}_1\mathrm{V}={\mathrm{\mu }}_1\mathrm{RT}$ and ${\mathrm{P}}_2\mathrm{V}={\mathrm{\mu }}_2\mathrm{RT}$ 

Here, $1$ and $2$ refer to neon gas and oxygen gas respectively

Now, $\mathrm{ }\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}=\frac{{\mathrm{\mu }}_1}{{\mathrm{\mu }}_2}\Rightarrow \frac{{\mathrm{\mu }}_1}{{\mathrm{\mu }}_2}=\frac{3}{2}$ ( given $\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}=\frac{3}{2}$)
If ${\mathrm{N}}_1$ and ${\mathrm{N}}_2$ are number of molecules of two gases, then
$\therefore \hspace{0.17em}\hspace{0.17em}\frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}=\frac{{\mathrm{\mu }}_1}{{\mathrm{\mu }}_2}=\frac{3}{2}$, where ${\mathrm{\mu }}_1=\frac{{\mathrm{N}}_1}{{\mathrm{N}}_{\mathrm{A}}}$ and ${\mathrm{\mu }}_2=\frac{{\mathrm{N}}_2}{{\mathrm{N}}_{\mathrm{A}}}$