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^ th floor. The number of…

Total number of persons = $9$9

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

Number of available floors = $8$8

The number of ways to select $4$4 persons out of $9$9 to form the first group is $^{9}C_{4}$9C4​. The remaining $5$5 persons automatically form the second group.

$$^{9}C_{4} = \dfrac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$$9C4​=4×3×2×19×8×7×6​=126

Now, these two distinct groups must exit at two different floors out of the $8$8 available floors. The number of ways to assign $2$2 different floors to the $2$2 groups is $^{8}P_{2}$8P2​.

$$^{8}P_{2} = 8 \times 7 = 56$$8P2​=8×7=56

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

Answer: $7056$7056