5.17 Let \(a < b < c < d\). If the variables \(x, y, z, t\) are some permutation of the numbers \(a, b, c, d\), how many different values can the expression
$$n=(x-y)^{2}+(y-z)^{2}+(z-t)^{2}+(t-x)^{2}$$
take?
[Solution] Let $n(x, y, z, t) = (x-y)^{2} + (y-z)^{2} + (z-t)^{2} + (t-x)^{2}$. Since the expression
$$\begin{aligned}
& n(x, y, z, t) + (x-z)^{2} + (y-z)^{2} \\
= & n(x, y, z, t) + \left(x^{2} + y^{2} + z^{2} + t^{2}\right) - 2(x z + y t)
\end{aligned}$$

is independent of the permutation of the numbers $a, b, c, d$ chosen as the values of the variables, and $x^{2} + y^{2} + z^{2} + t^{2}$ takes the same value for any permutation of the values $a, b, c, d$, the value of $n$ depends only on the value of $V = xz + yt$, and reaches its maximum and minimum values together with $V$. $\square$
$V$ can take three different values:
$$V_{1} = ab + cd, V_{2} = ac + bd, V_{3} = ad + bc.$$

And it satisfies the inequalities $V_{1} > V_{2} > V_{3}$, because
$$\begin{array}{l}
V_{1} - V_{2} = (d - a)(c - b) > 0 \\
V_{2} - V_{3} = (b - a)(d - c) > 0
\end{array}$$

Therefore, the values of $n$ can be arranged in size as follows:
$$n(a, c, b, d) > n(a, b, c, d) > n(a, b, d, c).$$