Explanation:

To find the maximum value of $n$ such that $5^n$ divides $30!+35!$:

- Factorial powers of 5:

- ${\nu }_5(30!)$ = $\lfloor \frac{30}{5}\rfloor +\lfloor \frac{30}{25}\rfloor =6+1=7$

- ${\nu }_5(35!)$ = $\lfloor \frac{35}{5}\rfloor +\lfloor \frac{35}{25}\rfloor =7+1=8$

- Expression simplification:

- $30!+35!=30!(1+31\times 32\times 33\times 34\times 35)$

- Since $31\times 32\times 33\times 34\times 35$ contains one 5 factor, $5$ divides the expression once.

- Conclusion:

- Minimum of ${\nu }_5(30!)$ and ${\nu }_5(35!)-1$ is $7$

- Correct Option:

- Option 3: 7