KarunMendiratta wrote:
OA Explanation
(5th 5th 5th) _ _ _ _ _ _ _
Since, there are 10 people and exactly 3 get down on 5th floor, there are 10C3 ways in which we can choose 3 people out of 10 for the same i.e. 120 ways
Now, rest 7 will have only 11 floor options, they can get down in 11 x 11 x 11 x 11 x 11 x 11 x 11 ways i.e. 117 ways
Col. A: 120 x 117
Col. B: 2.4 x 109
Col. A: 120 x 117
Col. B: 240 x 107
Dividing both sides by 120;
Col. A: 117
Col. B: 2 x 107
Dividing both sides by 107;
Col. A: (1110)7
Col. B: 2
Col. A: (1.10)7
Col. B: 2
Clearly, Col. A < Col. B
Hence, option B
I have a little uestion here:
why are we calculating 10C3 instead of 10P3?.
My reasoning is the following,
lets say that every person is labeled with a letter: (A,B,C..). Because order matters, we have to count the order in which the group get down from the lift
Person | floor
A | 1
B | 2
C | 3
D | 4
E | 5
F | 5
G | 5H | 8
J | 9
K | 10
in this case, person E, F and G got down on the 5th floor (in that order). However, we could also have this case:
Person | floor
A | 1
B | 2
C | 3
D | 4
F | 5
E | 5
G | 5H | 8
J | 9
K | 10
In this situation, person F got down first, followed by E and G, which is clearly a different combination, which in my opinion, is not captured by 10C3.
am I mistaken?,
Thanks!!