44. In the kitchen, there are five light bulbs. How many lighting options exist?

Do you understand what lighting options mean? Each lamp can be on or off. Two options are considered different if they differ in the state of at least one lamp (see Fig. 7 and 8).
44. First solution. The number of ways to illuminate with one lamp out of five is equal to the number of ways to illuminate with four lamps out of five (Fig. 32).

$$
\begin{aligned}
& 000 \sim 0000 \\
& 10000000 \\
& 0000 \text { } \\
& 1000 \\
& 0000 \sim 0000
\end{aligned}
$$

## Fig. 32

Similarly, the number of ways to illuminate with two lamps out of five is equal to the number of ways in which two lamps out of five do not light up (i.e., three lamps out of five light up). These methods are shown in Fig. 33 (10 ways in which two lamps light up and 10 ways in which three lamps light up).

## $00 \bullet \bullet \sim \bullet 000$ $0 \bullet 0 \cdot \sim \sim 000$ $0 \bullet 0 \cdot \sim 00 \bullet 0$ $0 \cdot \bullet \cdot 0 \sim 000$ $-00 \cdot \sim 0 \cdot 000$ $-0 \cdot 0 \sim 0 \cdot 00$ $-0 \cdot 0 \sim 000$ $\cdot 00 \cdot \sim 00 \cdot 0$ $\bullet 0.0 \sim 00 \cdot 0$ $\bullet \bullet 00 \sim 000 \bullet$

## Beo naunes ropar:

00000

All lamps in the roper: $-00$

Fig. 3 s
2 lamps, and 10 ways in which 3 lamps light up). There are also two more ways (Fig. 34) (all lamps light up, all lamps do not light up). We have considered all ways of illumination and it is now easy to count them. Second solution. 1) Suppose there is one lamp in the kitchen. It can be in two states: not lit or lit 0, i.e., there are 2 ways of illumination.
2) 2 lamps. The first lamp can be in two states: and $\bigcirc$.

Each of these can be combined with any of the two states of the second lamp: if the second lamp is not lit, we get 2 ways of illumination.

![](https://cdn.mathpix.com/cropped/2024_05_21_683be32ee1eca857285bg-093.jpg?height=122&width=149&top_left_y=1600&top_left_x=672)

If the second lamp is lit, we get 2 more ways

![](https://cdn.mathpix.com/cropped/2024_05_21_683be32ee1eca857285bg-094.jpg?height=127&width=151&top_left_y=204&top_left_x=366)

In total, there are 4 ways of illumination:

![](https://cdn.mathpix.com/cropped/2024_05_21_683be32ee1eca857285bg-094.jpg?height=116&width=366&top_left_y=445&top_left_x=251)
3) 3 lamps. The first two lamps can be in four states. Each of these four states can be combined with any of the two states of the third lamp.
![](https://cdn.mathpix.com/cropped/2024_05_21_683be32ee1eca857285bg-094.jpg?height=182&width=768&top_left_y=832&top_left_x=98)

In total, there are 8 ways of illumination:

![](https://cdn.mathpix.com/cropped/2024_05_21_683be32ee1eca857285bg-094.jpg?height=155&width=750&top_left_y=1164&top_left_x=89)
4) 4 lamps. The first three lamps can be in eight states. Each of these can be combined with any of the two states of the fourth lamp. There are $8 \cdot 2=16$ ways of illumination.
5) 5 lamps. There are $16 \cdot 2=32$ ways of illumination. The method of solution considered allows for minor modifications: the transition from two lamps to three
can be illustrated not only by the above scheme but also by the following schemes:
![](https://cdn.mathpix.com/cropped/2024_05_21_683be32ee1eca857285bg-095.jpg?height=196&width=824&top_left_y=262&top_left_x=159)

Both of them also show that adding one lamp (which can be in two states) doubles the number of ways of illumination.