Example 2. Use 6 white beads, 8 blue beads, and 1 red bead to form a string. How many different ways can this be done?

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Example 2. Use 6 white beads, 8 blue beads, and 1 red bead to form a string. How many different ways can this be done?
This is a circular permutation problem. If we fix the red bead, it turns into a linear permutation problem. By the principle of the whole, apart from the red bead, the remaining 14 beads have $N_{1}=\frac{P_{14}^{14}}{P_{6}^{6} \cdot P_{8}^{8}}=3003$ ways to be strung.

Below, we discuss two cases based on whether the 14 beads are symmetrical or asymmetrical relative to the red bead.
(1) The number of ways to string the beads symmetrically about the red bead is
$$
N_{2}=\frac{P_{7}^{7}}{P_{3}^{3} \cdot P_{4}^{4}}=35 \text { (ways). }
$$
(2) The number of ways to string the beads asymmetrically about the red bead is
$$
N^{\prime}=N_{1}-N_{2}=2963 \text { (ways). }
$$

If only beads of the same color are swapped, the resulting stringing methods are considered the same, so the number of asymmetrical stringing methods relative to the red bead is
$$
N_{2}=\frac{1}{2} N^{\prime}=1484 \text { (ways). }
$$

Therefore, the total number of different stringing methods is $N=N_{2}+1-N_{3}=1519$ (ways).

Note: When solving such problems, the principle of classification often starts from the process of the problem, while the principle of the whole often starts from the essence of the problem. They observe the problem from different angles. Therefore, when applying them, one should not emphasize one aspect while neglecting the other.