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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

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!

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! _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

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. _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

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.

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