Three persons enter in a lift at the ground floor. The lift will go upto $$10^{th}$$ floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to________

Three persons exit a lift at three different floors. The lift does not stop at floors 1, 2, or 3. Find the number of ways.

The lift goes from ground floor to 10th floor. It does not stop at 1st, 2nd, and 3rd floors. So the available exit floors are: 4, 5, 6, 7, 8, 9, 10 — a total of 7 floors.

The three persons must exit at three different floors chosen from these 7. The first person can choose any of 7 floors, the second can choose any of the remaining 6, and the third can choose any of the remaining 5:

$$\text{Number of ways} = P(7, 3) = 7 \times 6 \times 5 = 210$$

Alternatively, this is $$\frac{7!}{(7-3)!} = \frac{7!}{4!} = \frac{5040}{24} = 210$$.

The correct answer is 210.