In a classroom, 24 light fixtures were installed, each capable of holding 4 bulbs. When some of the fixtures were filled with four bulbs, it became clear that the available stock would be insufficient. Thereafter, they started to screw in the bulbs in groups of three, then in pairs, and finally individually. Unfortunately, even this way, there were still fixtures that did not receive any bulbs. How many bulbs were missing if twice as many fixtures received one bulb as those that received four, and half as many fixtures received no bulbs at all as those that received three?
I. solution. Each of the 24 lamp holders can fit 4 bulbs. Let's create a table:

$$
\begin{array}{lccccl}
\text { Number of screwed-in bulbs: } & 4 & 3 & 2 & 1 & 0 \\
\text { Number of lamp holders: } & x & 2 y & 24-3 x-3 y & 2 x & y
\end{array}
$$

Based on this, we finally get

$$
4 x+6 y+2(24-3 x-3 y)+2 x+0 \cdot y=4 x+6 y+48-6 x-6 y+2 x=48
$$

bulbs were screwed in.

Since a total of $24 \cdot 4=96$ bulbs could fit in the 24 lamp holders, $96-48=48$ bulbs were missing.

II. solution. We will show that by rearranging the bulbs, it is possible to have 2 bulbs in every lamp. This means that $24 \cdot 2=48$ bulbs were missing.

If 2 bulbs are removed from each lamp that has 4 bulbs and 1 bulb is placed in each of two lamps that previously had only 1 bulb, then every lamp that previously had 4 or 1 bulbs will now have 2 bulbs.

If 1 bulb is removed from each of two lamps that have 3 bulbs and placed in a lamp that had no bulbs at all, then every lamp will have 2 bulbs.