Three persons enter in a lift at the ground floor. The lift will go...

Three persons must exit at three different floors.
The lift goes to the 10th floor and does not stop at floors 1, 2, and 3.
Available floors for exit: 4, 5, 6, 7, 8, 9, 10 (7 floors total).
The number of ways to assign 3 persons to 3 different floors from 7 available floors is a permutation problem: $P(7, 3) = 7 \times 6 \times 5 = 210$.
Alternatively, choose 3 floors from 7 and arrange 3 persons: ${}^{7}P_{3} = 35 \times 6 = 210$.