Question

In a room, there are 12 bulbs of the same wattage, each having a separate switch. The number of ways to light the room with different amount of illumination is:
(a) ${{12}^{2}}-1$
(b) ${{2}^{12}}$
(c) ${{2}^{12}}-1$
(d) ${{12}^{2}}$

Answer 1

First of all, find the number of ways for each bulb, that is, if it is turned OFF or turned ON. Now, multiply the ways of all 12 bulbs to get the total ways. Also, subtract that way when they would be no illumination in the room.

Complete step-by-step answer:
In the question, we are given that in a room there are 12 bulbs of the same wattage, each having a separate switch. We have to find the number of ways to light the room with different amounts of illumination.
We know that there are two ways for every bulb that is each bulb can be either switched on or switched off. In other words, we can say that either a bulb can be illuminated or not illuminated at a particular time.
We are also given that there are a total of 12 bulbs in the room and all of the bulbs are of the same wattage and have separate switches. So, out of 12 bulbs, the first bulbs can be either switched on or off; that is a total 2 cases. Similarly, the second bulb can be either switched on or off, so a total of 2 cases. Similarly, the third bulb also can be either switched on or off, so a total of 2 cases.
Similarly, all the remaining bulbs could either be switched on or off. So, there would be 2 cases for each of the bulbs from the 4th to 12th bulb.
Now, since there are 2 cases for each bulb from 1st bulb to 12th bulb individually. So, we get,
The number of ways of selecting ON or OFF for all the 12 bulbs together

& =2\times 2\times 2\times 2.....12\text{ times} \\\ & \text{=}{{\text{2}}^{12}} \\\ \end{aligned}$$ In the above number of ways, that way would also be included when all the bulbs would be in the OFF position simultaneously. But we are given that, there must be some amount of illumination in the room. So, we have to exclude that case from the above case when all the bulbs would be in the OFF position. So, we get the number of ways to light the room with different amounts of illumination $$={{2}^{12}}-1$$. Hence, the option (c) is the right answer. **Note:** In this question, many students give the answer as $${{2}^{12}}$$ which is wrong. They give this answer because they often forget to consider the case when all the bulbs would be in the OFF position or there would be no illumination in the room. So, this must be taken care of. Also, some students give the answer as $${{12}^{2}}-1$$ thinking that 12 bulbs could either be turned ON or OFF, so, $$12\times 12={{12}^{2}}ways$$ which is wrong because each bulb can be turned ON or OFF individually because each has a separate switch, so total $${{2}^{12}}\text{ ways}$$.