Can a polyhedron have 10 faces,

No, because every polyhedron satisfies Euler's formula, given below:

$\mathrm{F}+\mathrm{V}=\mathrm{E}+2$

Here, number of faces $\mathrm{F}=10$

Number of edges $E=20$

Number of vertices $\mathrm{V}=15$

So, by Euler's formula:

LHS : $10+15=25$

RHS : $20+2=22$,

which is not true because $25 \neq 22$

Hence, Eulers formula is not satisfied and no polyhedron may be formed.