hall can be lighted

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srcc25anu: Master | Next Rank: 500 Posts; Posts: 423; Joined: Fri Jun 11, 2010 7:59 amFollowed by: 2 members; Location: Seattle, WA; Thanked: 86 times

by srcc25anu » Wed Mar 09, 2011 10:56 pm

There are 10 lamps in a hall.Each one of them can be switched independently.Then How many number of ways in which the hall can be illuminated?

A.2^(10) -1
B.2^(10)
C.2
D.2^(10) -2
E.None of these

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Source: Beat The GMAT — Problem Solving | Wed Mar 09, 2011 10:56 pm

shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Wed Mar 09, 2011 11:14 pm

srcc25anu wrote:There are 10 lamps in a hall.Each one of them can be switched independently.Then How many number of ways in which the hall can be illuminated?

A.2^(10) -1
B.2^(10)
C.2
D.2^(10) -2
E.None of these

10 lamps can be arranged in 2^10 ways.

We need at least one light to be switched at a time. That means we have to discard only one case where all the lights are turned off.

So IMO [spoiler]answer should be 2^10 - 1 (A)[/spoiler]

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manpsingh87: Master | Next Rank: 500 Posts; Posts: 436; Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members; Thanked: 72 times

by manpsingh87 » Thu Mar 10, 2011 1:27 am

srcc25anu wrote:There are 10 lamps in a hall.Each one of them can be switched independently.Then How many number of ways in which the hall can be illuminated?

A.2^(10) -1
B.2^(10)
C.2
D.2^(10) -2
E.None of these

As each bulb can change its state in two way (either ON/OFF) therefore, we can perform two operation with each bulb, now hall will be illuminated if at least one of the bulb is switched on. considering this situation we have;

2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10 = 2^10 - 1

as 2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10 = 2^10;

hence answer should be option A

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aleph777: Master | Next Rank: 500 Posts; Posts: 131; Joined: Fri Jun 18, 2010 10:19 am; Location: New York, NY; Thanked: 10 times

by aleph777 » Thu Mar 10, 2011 6:43 am

Just want to make sure I understand this correctly--it's essentially based on the fundamental counting principle, correct?

The total number of possible arrangements is simply the factor of each individual operation.

Since there are 10 lights, and each can be turned one or off, then there are 2 possible arrangements for each of the ten lights. Therefore:

2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 2^10. However, since we are solving for the number of possible combinations where lights are ON, we need to subtract the one possibility that lights are off. Therefore, 2^10 - 1.

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