MHB Solve Bracelet Problem w/14 Beads (Red, White, Blue)

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How many essentially different bracelets can be made from 14 beads consisting of 6 red beads, 5 white beads, and 3 blue beads? Arrangements obtained by rotation or reflection are considered equivalent.

I have being trying to use Burnside's lemma to solve this. My group of transformations has 28 elements - the identity, 13 rotations, 7 reflections along lines between 2 beads, and 7 reflections through 2 beads. Number the beads from 1 to 14. Here is an example of a reflection between 2 beads: (1 14)(2 13)(3 12)(4 11)(5 10)(6 9)(7 8). Here is an example of a reflection through 2 beads: (1)(2 14)(3 13)(4 12)(5 11)(6 10)(7 9)(8).

My solution is: $$\dfrac{\dfrac{14!}{6!5!3!} + 7 \cdot 120}{28}=6036$$.

Can anyone confirm this or point out my error or provide a better strategy?

Thanks
 
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I'll confirm it.
And no, I'm not aware of a better strategy.
 
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