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Worked Examples · Example 35

Q.How many necklaces can be made using 20 beads, 8 being blue, 5 green, 5 yellow and 2 red.

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20 beads (8 blue, 5 green, 5 yellow, 2 red) strung into a necklace: 19!2×8! 5! 5! 2!=52,378,326\dfrac{19!}{2\times8!\,5!\,5!\,2!}=52{,}378{,}326 distinct necklaces.

For beads with repeated colours, circular (rotation-only) arrangements =(n−1)!p1! p2! ⋯=\dfrac{(n-1)!}{p_1!\,p_2!\,\cdots}. A necklace further allows flipping (reflection), so divide by one extra factor of 22:

(n−1)!2 p1! p2! ⋯\dfrac{(n-1)!}{2\,p_1!\,p_2!\,\cdots}

  1. Total beads n=20n=20: blue p1=8p_1=8, green p2=5p_2=5, yellow p3=5p_3=5, red p4=2p_4=2 (check: 8+5+5+2=208+5+5+2=20 ✓).
  2. Linear arrangements (if laid in a row) =20!8! 5! 5! 2!=\dfrac{20!}{8!\,5!\,5!\,2!}.
  3. For a bracelet (circular, rotation-only), divide by n=20n=20:

20!20×8! 5! 5! 2!=19!8! 5! 5! 2!\dfrac{20!}{20\times8!\,5!\,5!\,2!}=\dfrac{19!}{8!\,5!\,5!\,2!}

  1. Compute: 19!=121,645,100,408,832,00019!=121{,}645{,}100{,}408{,}832{,}000 and 8! 5! 5! 2!=40320×120×120×2=1,161,216,0008!\,5!\,5!\,2!=40320\times120\times120\times2=1{,}161{,}216{,}000. 19!8! 5! 5! 2!=121,645,100,408,832,0001,161,216,000=104,756,652\dfrac{19!}{8!\,5!\,5!\,2!}=\dfrac{121{,}645{,}100{,}408{,}832{,}000}{1{,}161{,}216{,}000}=104{,}756{,}652 …

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