5. A subset $M$ of the set $S=\{1928,1929, \cdots, 1949\}$ is called "red" if and only if the sum of any two distinct elements in $M$ is not divisible by 4. Let $x$ and $y$ represent the number of red four-element subsets and red five-element subsets of $S$, respectively. Try to compare the sizes of $x$ and $y$, and explain the reason.

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5. Divide the elements of set $S$ into residue classes modulo 4 as $S_{0}, S_{1}, S_{2}, S_{3}$, where the elements of $S_{i} (i=0,1,2,3)$ have a remainder of $i$ when divided by 4.
It is easy to see that $\left|S_{0}\right|=\left|S_{1}\right|=6,\left|S_{2}\right|=\left|S_{3}\right|=5$.
For any red subset $M$ of set $S$, we have
$$
\left|M \cap S_{0}\right| \leqslant 1,\left|M \cap S_{2}\right| \leqslant 1,
$$

and $M$ cannot contain elements from both $S_{1}$ and $S_{3}$ (otherwise, $M$ will have two numbers whose sum is divisible by 4, which contradicts the definition of a red subset). Conversely, when the above conditions are met, $M$ must be a red subset.
First, find the number of ways to choose a red four-element subset $M$ of $S$, denoted as $x$.
When $\left|M \cap S_{0}\right|=\left|M \cap S_{2}\right|=0$, $M$ is a subset of $S_{1}$ or a subset of $S_{3}$, so there are
$$
\mathrm{C}_{6}^{4}+\mathrm{C}_{5}^{4}=20
$$

ways to choose $M$.
When $\left|M \cap S_{0}\right|=1,\left|M \cap S_{2}\right|=0$ or $\left|M \cap S_{0}\right|=$ $0,\left|M \cap S_{2}\right|=1$, $M$ has three elements that are simultaneously in $S_{1}$ or simultaneously in $S_{3}$, so there are
$$
\left(\left|S_{0}\right|+\left|S_{2}\right|\right)\left(\mathrm{C}_{6}^{3}+\mathrm{C}_{5}^{3}\right)=330
$$

ways to choose $M$.
When $\left|M \cap S_{0}\right|=\left|M \cap S_{2}\right|=1$, $M$ has two elements that are simultaneously in $S_{1}$ or simultaneously in $S_{3}$, so there are
$$
\left|S_{0}\right|\left|S_{2}\right|\left(\mathrm{C}_{6}^{2}+\mathrm{C}_{5}^{2}\right)=750
$$

ways to choose $M$.
In summary, $x=20+330+750=1100$.
Similarly, the number of ways to choose a red five-element subset $M$ of $S$ is
$$
\begin{aligned}
y= & \left(\mathrm{C}_{6}^{5}+\mathrm{C}_{5}^{5}\right)+\left(\left|S_{0}\right|+\left|S_{2}\right|\right)\left(\mathrm{C}_{6}^{4}+\mathrm{C}_{5}^{4}\right)+ \\
& \left|S_{0}\right|\left|S_{2}\right|\left(\mathrm{C}_{6}^{3}+\mathrm{C}_{5}^{3}\right) \\
= & 7+220+900=1127 .
\end{aligned}
$$

Therefore, $x<y$.