Question ID: #973
Three persons enter in a lift at the ground floor. The lift will go upto $10^{\text{th}}$ 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 ________.
Solution:
The lift goes from the ground floor up to the $10^{\text{th}}$ floor.
Available floors for exiting are: $4^{\text{th}}, 5^{\text{th}}, 6^{\text{th}}, 7^{\text{th}}, 8^{\text{th}}, 9^{\text{th}}, 10^{\text{th}}$.
Total number of available floors $n = 7$.
We need to select 3 different floors out of these 7 floors for the 3 persons to exit.
Number of ways to select 3 different floors:
$${}^nC_r = {}^7C_3 = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$$
Once the 3 floors are selected, the 3 distinct persons can exit on these floors in $3!$ ways.
Number of ways to arrange the persons:
$$3! = 3 \times 2 \times 1 = 6$$
Total number of ways is the product of selection and arrangement:
$$\text{Total ways} = {}^7C_3 \times 3!$$
$$\text{Total ways} = 35 \times 6 = 210$$
Ans. 210
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