Question

There are 10 lamps in a hall. Each one of them can be switched on independently. The number of ways in which the hall can be illuminated is
A. ${10^2}$
B. $1023$
C. ${2^{10}}$
D. $10!$

Answer 1

Hint: First, list all the possible ways that can be made from these 10 lamps in a hall. There will be the only way where the hall cannot be illuminated. As the hall is to be illuminated, then subtract the way when the hall cannot be illuminated from all possible ways. So, the remaining value will give the number of ways in which the hall can be illuminated.

Complete step-by-step solution:
Given: - There are 10 lamps in a hall. Each one of them can be switched on independently.
As every bulb has 2 options either on or off and there is a total of 10 bulbs in the hall. So, the total number of ways will be,
$ \Rightarrow {2^{10}} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
Multiply the terms on the right side,
$ \Rightarrow {2^{10}} = 1024$
There will be only one way when the hall will not be illuminated.
Now, subtract the one way from the total number of ways to get the number of ways in which the hall can be illuminated,
$ \Rightarrow {2^{10}} - 1 = 1024 - 1$
Subtract the terms,
$ \Rightarrow {2^{10}} - 1 = 1023$
Thus, the number of ways in which the hall can be illuminated is 1023.

Hence, option (B) is the correct option.

Note: The fundamental principle of counting is used to find the total ways.
The fundamental counting principle is a rule used to count the total number of possible outcomes in a situation. It states that if there are $n$ ways of doing something, and $m$ ways of doing another thing after that, then there are $n \times m$ ways to perform both of these actions. In other words, when choosing an option for $n$ and an option for $m$, there are $n \times m$ different ways to do both actions.