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There are 12 floors in a building and 10 persons are about to leave

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KarunMendiratta
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There are 12 floors in a building and 10 persons are about to leave [#permalink] New post   06 Mar 2021, 04:42
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There are 12 floors in a building and 10 persons are about to leave a lift. Exactly 3 among them would leave at 5th floor.

Quantity A
Quantity B
Number of ways they can leave the lift
\(2.4\) x \(10^9\)


(A) Quantity A is greater
(B) Quantity B is greater
(C) The two quantities are equal
(D) The relationship cannot be determined from the information given
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B
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There are 12 floors in a building and 10 persons are about to leave [#permalink] New post   06 Mar 2021, 23:40
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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
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Re: There are 12 floors in a building and 10 persons are about to leave [#permalink] New post   11 Mar 2021, 02:31
question is hard or am i stupid?
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Re: There are 12 floors in a building and 10 persons are about to leave [#permalink] New post   11 Mar 2021, 02:45
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void wrote:
question is hard or am i stupid?


It is hand. On GRE it would be somewhere around 165+
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There are 12 floors in a building and 10 persons are about to leave [#permalink] New post   16 Mar 2021, 12:48
KarunMendiratta wrote:
OA Explanation

(5th 5th 5th) _ _ _ _ _ _ _

Since, there are \(10\) people and exactly \(3\) get down on 5th floor, there are \(^{10}C_2\) 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


Quote:
Since, there are 10 people and exactly 3 get down on 5th floor, there are 10C2 ways in which we can choose 3 people out of 10 for the same i.e. 120 ways


why it is not 10C3? I don't understand it.
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There are 12 floors in a building and 10 persons are about to leave [#permalink] New post   16 Mar 2021, 20:26
nurakib wrote:
KarunMendiratta wrote:
OA Explanation

(5th 5th 5th) _ _ _ _ _ _ _

Since, there are \(10\) people and exactly \(3\) get down on 5th floor, there are \(^{10}C_2\) 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


Quote:
Since, there are 10 people and exactly 3 get down on 5th floor, there are 10C2 ways in which we can choose 3 people out of 10 for the same i.e. 120 ways


why it is not 10C3? I don't understand it.


It is \(^{10}C_3\) only, I have edited the explanation. Thanks!
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Re: There are 12 floors in a building and 10 persons are about to leave [#permalink] New post   22 Sep 2022, 12:00
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 | 5
H | 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 | 5
H | 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!!
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Re: There are 12 floors in a building and 10 persons are about to leave [#permalink] New post   07 Dec 2023, 22:56
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