A necklace is made out of identically shaped, coloured beads. Ten beads are red, 9 are black, 7 are white, 12?

Question

A necklace is made out of identically shaped, coloured beads. Ten beads are red, 9 are black, 7 are white, 12 are pink, and 4 are purple. The beads are strung together on a waxed string. The ends are tied together to form a large knot that prevents the beads from sliding past it. How many different necklaces can be made in this way?

Answer 1

Since the knot is there, you get to work with linear permutations (not cyclic ones). So, really, you want to know how many orders you can make with the beads. The tricky part is that, with a necklace, you can flip it over. So if the two orders may actually make the same necklace. The unfortunate part is that this may only be true SOMETIMES... if the necklace is symmetric, then flipping it over doesn't change it.

The general theorem here is know as "Burnside's Lemma". You can check it out here:

http://en.wikipedia.org/wiki/Burnside's_lemma

For your problem, the group is the two-element group of flipping and the elements fixed by the non-identity element of the group are the "palindromic" patterns (same backwards and forwards) and (trivially) the identity element fixes everything.

To figure out who many total patterns there are, you have (let (x C y) mean (x choose y))

(42 C10)*(32 C 9)*(23 C 7)*(16 C 12)*(4 C 4).

Now you have to figure out how many are palindromic. But you are lucky - since there are an even number of beads, in order for a pattern to be palindromic, it must have all it's color have an even number. So none of the patterns here can be palindromic.

Thus, by Burnside's Lemma (which is trivial now, since it is just noticing that each necklace forms 2 patterns distinct patterns and so you are overcounting by a factor of 2), the total number of necklaces is:

(1/2)*(42 C10)*(32 C 9)*(23 C 7)*(16 C 12)*(4 C 4)

Anyway, it's a huge number, so I'll leave it like that (if you want to calculate it, be my guest).

Answer 2

probability of 18 for each place on the 12-beaded necklace in case you enable repetition; which would be 18^12 (18 raised to the 12th means) or a million.157 x 10^15 mixtures achievable 18 x 18 x 18 x ...and so on for 12 positions now in case you would be unable to repeat colorations, then 18 x 17 x sixteen x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 or 8.892 x 10^12 mixtures

Answer 3

one

Answer 4

fffffff ooooooo ooooooo lllllll

Answer 5

solve it using circular permutation.....

Answer 6

544786209316058000000000000.

Answer 7

158,462,223,897,460