circular permutation in case of identical diamond and pearls

Question

the number of ways a necklace can be formed by 18 identical diamond and 3 identical pearl ? my solution is that
divide the procedure into three case case1:all pearls together that leads to only one case. case 2: two pearls together and one seperate that leads to 9 cases and last case 3 all seperate :which i am unable to do plzz help and

ANWSER IS 37

Are you familiar with multinomials? – Andres Mejia Mar 20 '16 at 18:56 — Andres Mejia, Mar 20 '16 at 18:56
yes little bit you can try answering me – kola Mar 20 '16 at 19:13 — kola, Mar 20 '16 at 19:13

score 0 · Accepted Answer · answered Mar 20 '16 at 20:51

0

You have made a good start. Now you need to list the possibilities for the number of diamonds in the three groups in case 3. You need three positive integers totalling 18. So you have: 18=16+1+1, 15+2+1, 14+3+1, 14+2+2, 13+4+1, 13+3+2, 12+5+1, 12+4+2, 12+3+3 and so on down through 9+8+1 ... 8+8+2 ... 7+7+4 ... 6+6+6. You will find there are 27 of them. Each one gives just one necklace. Add them to the 10 you have already.

answered Mar 20 '16 at 20:51

almagest

18,380

is there any other way instead of counting 27 possibilities – kola Mar 20 '16 at 21:02

circular permutation in case of identical diamond and pearls

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