We need the number of electrons present in the $4d$ orbitals of:

• Niobium $(\mathrm{Nb})$
• Ruthenium $(\mathrm{Ru})$

Write their atomic numbers:

• $\mathrm{Nb}: Z=41$
• $\mathrm{Ru}: Z=44$

Find their ground-state electronic configurations.

For niobium $(\mathrm{Nb})$: The expected filling is near $[\mathrm{Kr}]\,5s^2 4d^3$, but due to stability, the actual configuration is $\mathrm{Nb} = [\mathrm{Kr}]\,4d^4 5s^1$ Hence, the number of electrons in $4d$ orbitals is $x=4$

For ruthenium $(\mathrm{Ru})$: Its actual ground-state configuration is $\mathrm{Ru} = [\mathrm{Kr}]\,4d^7 5s^1$ Hence, the number of electrons in $4d$ orbitals is $y=7$

Therefore, $x+y=4+7=11$

Final answer: $\boxed{11}$

Comparison with stored correct answer: Stored correct answer = $11$. Our derived answer also = $11$. Therefore, they agree.