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Given 10 beads on a necklace, 6 white and 4 red, how many

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Given 10 beads on a necklace, 6 white and 4 red, how many [#permalink] New post 25 Nov 2003, 01:43
Given 10 beads on a necklace, 6 white and 4 red, how many ways
canthe beads be arranged so that no three red beads are together?
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 [#permalink] New post 25 Nov 2003, 03:09
in my opinion:

total ways = 10!
forbidden ways = unite three red balls in one black: 4C3 ways=4
thus, we have 1 black, 1 red, and 6 white balls; there are 8!/6!*1!*1! ways to do so; or 7*8=56

10!-4*56=10!-224
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 [#permalink] New post 25 Nov 2003, 06:49
Let's calculate cases that undergo the limitation: let's assume that 1st, 2nd and 3rd places are taken by red bean. Then if either 4th or 10th place is taken by red bean as well would mean that 4 red beans are together - which is to me also under the limitation. Thus I would solve it this way:

total = 10!
possible combinations of red beans in a row = 4!
possible combinations of white beans when 4 red are in a row = 6!

outcome:
10! - (6!*4!)
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 [#permalink] New post 25 Nov 2003, 15:27
stolyar wrote:
in my opinion:

total ways = 10!
forbidden ways = unite three red balls in one black: 4C3 ways=4
thus, we have 1 black, 1 red, and 6 white balls; there are 8!/6!*1!*1! ways to do so; or 7*8=56

10!-4*56=10!-224


be careful. this problem has a lot of hidden complexities if properly solved. 1) a necklace can be rotated such that a number of arrangements are henceforth "indistinguishable". 2) a necklace with beads can also be "flipped over" (mirror image) adding yet another "axis of symmetry" -- these are hard to count because we need to exclude arrangement that are already symmetrical in order to avoid double counting.

Consider this: suppose we have a bracelet with 3 different colored beads (assume that there is no distiguishing clasp and the beads are symmetricallyl distributed around the bracelet). There are theoretically 3! or 6 different ways to arrange the beads. Suppose you had six bracelets each with one of those arrangements, then threw them all into a hat and mixed them up. Now dump them out of the hat and try to distinguish one for the other. You will realize that they are ALL the same arrangement!
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 [#permalink] New post 25 Nov 2003, 21:02
Akamai,

Side question; Isn't the total number of arrangements of objects (beads or whatever) on a circular ring calculated by the formula (n-1)! ? Thanks.
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 [#permalink] New post 25 Nov 2003, 22:23
I assumed that a necklace is a string, not a circle.
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 [#permalink] New post 26 Nov 2003, 09:00
stolyar wrote:
I assumed that a necklace is a string, not a circle.


I don't know how reasonable that assumption is (never seen anyone wear a necklace as a string) but even if that were true, you still missed the mirror image symmetry.
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 [#permalink] New post 18 Dec 2003, 20:10
I thought the number of combinations are ( assuming the necklace is a string instead of a circular necklace )

[ 10! / ( 6! * 4! ) ] - 4 * [ 8! / 6! * 1! * 1! ]

Can anyone explain me why this formula is wrong ( if it is wrong )
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 [#permalink] New post 24 Dec 2003, 17:34
I agree with anandnk. But have a small addition

[10!/(6!*4!)] - 4*[8!/(6!*2!)]

The reason for the extra 2! is that even though we take 3 red together, we will have 1 more red bead. And the (group or 3 red) + 1 more read make 2 simillar items.

Take also makes me think that we might not have counted the cases correctly.
Consider this: Rg Rg Rg are the three Red beads taken together. &
Ra is the 4th red bead.

say two Rg's are together (Rg, Rg) and Ra is next to it, this will make 3 red together, though we don't have all Rg's toether. Reason is that we have 4 reds and we are considering only 3 together and the 4th is in don't care position. I feel we can't do this. The 4th red bead also has to be accounted for.

OR may be my head is twisted too bad.
  [#permalink] 24 Dec 2003, 17:34
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Given 10 beads on a necklace, 6 white and 4 red, how many

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