KarunMendiratta wrote:
OA Explanation
(5th 5th 5th) _ _ _ _ _ _ _
Since, there are \(10\) people and exactly \(3\) get down on 5th floor, there are \(^{10}C_3\) 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. \(11^7\) ways
Col. A: \(120\) x \(11^7\)
Col. B: \(2.4\) x \(10^9\)
Col. A: \(120\) x \(11^7\)
Col. B: \(240\) x \(10^7\)
Dividing both sides by \(120\);
Col. A: \(11^7\)
Col. B: \(2\) x \(10^7\)
Dividing both sides by \(10^7\);
Col. A: \((\frac{11}{10})^7\)
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!!