14. Use 6 white beads, 8 black beads, and 1 red bead to string into a necklace. How many different ways are there to do this?
14. This is a circular permutation problem. If the red bead is fixed, it turns into a linear permutation problem. By the overall principle, apart from the red bead, the remaining 14 beads have $N_{1}=\frac{P_{14}^{14}}{P_{6}^{6} P_{8}^{8}}=3003$ ways of arrangement.
The 14 beads can be symmetric or asymmetric relative to the red bead in two scenarios:
（1）The number of symmetric arrangements relative to the red bead is $N_{2}=\frac{P_{7}^{7}}{P_{3}^{3} P_{4}^{4}}=35$.
（2）The number of asymmetric arrangements relative to the red bead is $N_{3}=N_{1}-N_{2}=2968$.
Since the same color beads can be swapped without changing the arrangement, the number of asymmetric arrangements relative to the red bead is $N_{4}=\frac{1}{2} N_{3}=1484$.
Therefore, the total number of different arrangements is $N=N_{2}+N_{3}=1519$.