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hall can be lighted

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hall can be lighted

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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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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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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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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