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m10#15

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Manager
Joined: 23 Jan 2011
Posts: 128

Kudos [?]: 110 [0], given: 13

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18 Jul 2011, 10:02
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.)

5
6
7
8
9

I do not understand the OE. Can some one explain this pls?

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.

Kudos [?]: 110 [0], given: 13

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1970

Kudos [?]: 2013 [0], given: 376

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18 Jul 2011, 10:18
Chetangupta wrote:
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.)

5
6
7
8
9

I do not understand the OE. Can some one explain this pls?

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.

Illuminated means at least one bulb is ON.

If
0 is OFF
1 is ON

000->Count this out because it's 3 OFFs
001
010
011
100
101
110
111

Total=7

Or:
Simply: 2^3-1=8-1=7
_________________

Kudos [?]: 2013 [0], given: 376

Current Student
Joined: 26 May 2005
Posts: 555

Kudos [?]: 234 [0], given: 13

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18 Jul 2011, 10:27
Chetangupta wrote:
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.)

5
6
7
8
9

I do not understand the OE. Can some one explain this pls?

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.

there are only 2 option s = on/off = 2ways
now if u try this on 3lamps = 2^3 ways = 8 ways

on on on
on on off
on off off........................... this is not valid ( atleast one should be on)
on off on
off off off
off on on
off off on
off on off

Kudos [?]: 234 [0], given: 13

Re: m10#15   [#permalink] 18 Jul 2011, 10:27
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m10#15

Moderator: Bunuel

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