A building has ground floor and 10 more floors. Nine persons enter in a lift at the ground floor. The lift goes up to the $10^{\text{th}}$ floor. The number of ways, in which any 4 persons exit at a floor and the remaining 5 persons exit at a different floor, if the lift does not stop at the first and the second floors, is equal to :

Total number of persons = $9$

The lift goes up to the $10^{\text{th}}$ floor but does not stop at the $1^{\text{st}}$ and $2^{\text{nd}}$ floors. Thus, the available floors for the persons to exit are from the $3^{\text{rd}}$ to the $10^{\text{th}}$ floor.

Number of available floors = $8$

The number of ways to select $4$ persons out of $9$ to form the first group is $^{9}C_{4}$. The remaining $5$ persons automatically form the second group.

$^{9}C_{4} = \dfrac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$

Now, these two distinct groups must exit at two different floors out of the $8$ available floors. The number of ways to assign $2$ different floors to the $2$ groups is $^{8}P_{2}$.

$^{8}P_{2} = 8 \times 7 = 56$

Total number of ways = $126 \times 56 = 7056$

Answer: $7056$