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In an auditorium, there are ten light switches each of which control

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In an auditorium, there are ten light switches each of which control  [#permalink]

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New post 03 Apr 2017, 00:30
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In an auditorium, there are ten light switches each of which control the lights of a different zone of the room. If there are only these 10 zones in the room and each light may only be set to "on" or "off", then how many different lighting arrangements are possible in the auditorium?

A. 10^10
B. 10!
C. 10^2
D 5!
E. 2^10

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Re: In an auditorium, there are ten light switches each of which control  [#permalink]

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New post 03 Apr 2017, 05:23
10 sets of ligthing, either on/off - 2 options only ... Hence

10 * 2C1 = 2^10 (E)
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Re: In an auditorium, there are ten light switches each of which control  [#permalink]

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New post 03 Apr 2017, 08:06
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Bunuel wrote:
In an auditorium, there are ten light switches each of which control the lights of a different zone of the room. If there are only these 10 zones in the room and each light may only be set to "on" or "off", then how many different lighting arrangements are possible in the auditorium?

A. 10^10
B. 10!
C. 10^2
D 5!
E. 2^10


Let the 10 light switches be Switch #1, Switch #2, Switch #3, etc.

Take the task of creating a lighting arrangement and break it into stages.

Stage 1: Set Switch #1
The switch can be ON or OFF. So, we can complete stage 1 in 2 ways

Stage 2: Set Switch #2
The switch can be ON or OFF. So, we can complete stage 2 in 2 ways

Stage 3: Set Switch #3
The switch can be ON or OFF. So, we can complete stage 3 in 2 ways

Stage 4: Set Switch #4
The switch can be ON or OFF. So, we can complete stage 4 in 2 ways
.
.
.

Stage 9: Set Switch #9
The switch can be ON or OFF. So, we can complete stage 9 in 2 ways

Stage 10: Set Switch #10
The switch can be ON or OFF. So, we can complete stage 10 in 2 ways

By the Fundamental Counting Principle (FCP), we can complete all 10 stages (and thus create a lighting arrangement) in (2)(2)(2)(2)(2)(2)(2)(2)(2)(2) ways (= \(2^{10}\) ways)

Answer:

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn the technique.

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Re: In an auditorium, there are ten light switches each of which control  [#permalink]

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New post 06 Apr 2017, 09:24
Bunuel wrote:
In an auditorium, there are ten light switches each of which control the lights of a different zone of the room. If there are only these 10 zones in the room and each light may only be set to "on" or "off", then how many different lighting arrangements are possible in the auditorium?

A. 10^10
B. 10!
C. 10^2
D 5!
E. 2^10


There are 10 light switches, and since each light switch has two possible options, the number of different possible lighting arrangements is 2^10.

Answer: E
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Re: In an auditorium, there are ten light switches each of which control  [#permalink]

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New post 08 Apr 2017, 18:30
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Hi All,

The 'math' behind this question isn't too difficult - once you recognize the 'pattern' involved. If you don't immediately see that pattern though, then you can still get to the correct answer by 'playing around' with simpler examples - and looking for the pattern. For example...

If there was just one light switch, then there would be TWO options: Off and On

If there were two light switches, then there would be FOUR options (I'll refer to Off as "F" and On as "N"):
FF
FN
NF
NN

If there were three light switches, then there would be EIGHT options:
FFF
FFN
FNF
NFF

NNN
NNF
NFN
FNN

Notice the pattern: 2...4...8... It certainly appears that every time we add another light switch, then number of possibilities DOUBLES. Thinking in these terms, we're looking for an answer that is based around lots of 'doubling'... and there's only one answer that exclusively multiplies a bunch of 2s together....

Final Answer:

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Re: In an auditorium, there are ten light switches each of which control  [#permalink]

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New post 08 Apr 2017, 21:00
Bunuel wrote:
In an auditorium, there are ten light switches each of which control the lights of a different zone of the room. If there are only these 10 zones in the room and each light may only be set to "on" or "off", then how many different lighting arrangements are possible in the auditorium?

A. 10^10
B. 10!
C. 10^2
D 5!
E. 2^10


For first switch total position= 2 (on or off)
For Second switch total position= 2 (on or off)
For Third switch total position= 2 (on or off)
For forth switch total position= 2 (on or off)... and so on...

Total arrangement = 2*2*2... 10 times = 2^10

Answer: Option E
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Re: In an auditorium, there are ten light switches each of which control  [#permalink]

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New post 24 Aug 2018, 07:17
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Hi, why do we not arrange the switches for each of the zones i.e. 10! way of arranging each of the switches for each of the zones and then try to either on/off the switch.
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Re: In an auditorium, there are ten light switches each of which control  [#permalink]

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New post 19 Oct 2018, 13:02
I have the same doubt
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Re: In an auditorium, there are ten light switches each of which control  [#permalink]

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New post 19 Oct 2018, 18:16
1
Hi anse and Mridul94,

The 'math' behind this question isn't too difficult - once you recognize the 'pattern' involved. If you don't immediately see that pattern though, then you can still get to the correct answer by 'playing around' with simpler examples - and looking for the pattern. For example...

If there were three light switches, then would the total number of possibilities be 3! = 6? Try 'mapping' out the options and see. The possible arrangements would be:

FFF
FFN
FNF
NFF

NNN
NNF
NFN
FNN

That's 8 arrangements - not 6, so we are clearly NOT dealing with a factorial. If you do similar work with 1 switch (2 arraignments) and 2 switches (4 arrangements), it certainly appears that every time we add another light switch, then number of possibilities DOUBLES. Thinking in these terms, we're looking for an answer that is based around lots of 'doubling'... and there's only one answer that exclusively multiplies a bunch of 2s together....

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Re: In an auditorium, there are ten light switches each of which control   [#permalink] 19 Oct 2018, 18:16
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