Three persons enter in a lift at the ground floor.

Question

Three persons enter in a lift at the ground floor. The lift will go up to 10th 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, i...

Answer: 210

Accepted Answer

Given:

The lift goes up to the ${10}^{}$ floor.

The lift does not stop at ${1}^{},{2}^{},{3}^{}$ floors.

So, the persons can get down only at the remaining floors.

$4,5,6,7,8,9,10\Rightarrow 7$

Now, all three persons must exit at three different floors (no two can get down at the same floor).

Also, the three persons are distinct, so who gets down where matters.

First, choose which 3 floors (out of 7) will be used:

${}^{7}{C}_{3}=\frac{7\cdot 6\cdot 5}{3\cdot 2\cdot 1}=35$

Next, assign (arrange) the 3 distinct persons to these 3 chosen floors:

Number of arrangements of 3 persons:

$3!=6$

So, total number of ways:

$={}^{7}{C}_{3}	imes 3!=35	imes 6=210$

Solutions · 1