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# M10-15

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Math Expert
Joined: 02 Sep 2009
Posts: 43896

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15 Sep 2014, 23:42
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61% (00:42) correct 39% (00:37) wrong based on 115 sessions

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There are three lamps in a hall. If each lamp can be switched on and off independently, in how many ways can the hall be illuminated? (The hall is illuminated when at least one of the lamps is on.)

A. 5
B. 6
C. 7
D. 8
E. 9
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 43896

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15 Sep 2014, 23:42
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Official Solution:

There are three lamps in a hall. If each lamp can be switched on and off independently, in how many ways can the hall be illuminated? (The hall is illuminated when at least one of the lamps is on.)

A. 5
B. 6
C. 7
D. 8
E. 9

As each lamp can be in either of the two modes (ON/OFF), there are $$2*2*2 = 8$$ possibilities. But as the stem says, the possibility when all the lamps are off does not count. Therefore the answer is 7.

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Intern
Joined: 22 Aug 2014
Posts: 47

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06 May 2016, 22:28
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If one lamp illuminates:3c1=3
If two lamps :3c2=3
If three lamps at a time=3c3=1
3+3+1=7

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Manager
Joined: 25 Nov 2009
Posts: 57
Location: India

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07 May 2016, 01:36
When A is on there are 4 possibilities: A, AB, AC, ABC
When B is on there are 2 new possibilities: B, BC (since we have already considered AB in the previous one)
When C is there's only 1 new possibility: C (since we have already considered AC & BC in the previous ones)
Thus in total there are 7 possibilities.
Intern
Joined: 25 Mar 2013
Posts: 7
Location: India
GMAT 1: 700 Q50 V39
WE: Information Technology (Consulting)

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17 Sep 2016, 07:50
I thought it like this.

Suppose 1st if off then total no of ways to 2nd the room:
1*2*2. = 4.

Similarly if 2nd or 3rd is off.

So total no of ways : 4+4+4 = 12.

Although, this option is not there. Please correct me where did I go wrong .
Math Expert
Joined: 02 Aug 2009
Posts: 5662

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17 Sep 2016, 09:04
Vishal Kumar wrote:
I thought it like this.

Suppose 1st if off then total no of ways to 2nd the room:
1*2*2. = 4.

Similarly if 2nd or 3rd is off.

So total no of ways : 4+4+4 = 12.

Although, this option is not there. Please correct me where did I go wrong .

Hi,
1*2*2...
The TWO in 1*2*2 includes when all three will be off and at that the room will not be lighted..
You are approaching the Q in a wrong way...
Room can be lighted in 3 ways..
1) all three on.. One way
2) two lights on.. choose two out of three 3C2...3 ways..
3) only one light on.. 3 ways
Total 1+3+3=7..

The way you are approaching, you could do it the following way..
The switch can be in two positions ON or OFF..
So total ways are 2*2*2=8, but these include the one instance when all three are OFF and at this time the room will not be lighted..
So 2*2*2-1=7
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Intern
Joined: 26 Jul 2016
Posts: 29
Location: India
Concentration: Operations, Entrepreneurship
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26 Oct 2016, 08:59
what am i missing here?

way that hall will be lit

3(if all 3 are on)x2(if two are on)x1(if one is on)
Intern
Joined: 10 May 2017
Posts: 27

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18 May 2017, 03:45
I assumed this problem in binary method.
001 to 111.
1 ~ 7.
Retired Moderator
Joined: 26 Nov 2012
Posts: 597

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29 Sep 2017, 03:40
Since in the question it is given that if one light is ON then the room illuminates. So here we are asked to find if at least one is ON.

We can consider the following cases.
1. Out of three any one light is ON - Room illuminates. OR
2. Any two out of three are ON - - Room illuminates. OR
3. All three are ON - - Room illuminates.

We can write this in equation form : 3c1 + 3c2 + 3c3 => 3 + 3 + 1 = > 7 ways.

Bunuel, is my process correct...please confirm..
Math Expert
Joined: 02 Sep 2009
Posts: 43896

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29 Sep 2017, 03:45
msk0657 wrote:
Since in the question it is given that if one light is ON then the room illuminates. So here we are asked to find if at least one is ON.

We can consider the following cases.
1. Out of three any one light is ON - Room illuminates. OR
2. Any two out of three are ON - - Room illuminates. OR
3. All three are ON - - Room illuminates.

We can write this in equation form : 3c1 + 3c2 + 3c3 => 3 + 3 + 1 = > 7 ways.

Bunuel, is my process correct...please confirm..

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Yes, that's correct.
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Re: M10-15   [#permalink] 29 Sep 2017, 03:45
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# M10-15

Moderators: chetan2u, Bunuel

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