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# One black, one red, one blue and two white pearls are

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Manager
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One black, one red, one blue and two white pearls are [#permalink]  07 Nov 2004, 14:14
One black, one red, one blue and two white pearls are attached to make a necklace. What is the probability that the black and the blue pearls are next to each other?
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Got 1/2 here
Total possible outcome: (n-1)! / 2 --> Because we are talking about beads on a necklace, we can just "flip" it over and we have the exact same reverse pattern so we have to eliminate those "mirror" patterns. 4!/2 = 12
You then have to divide it further by 2 because there are 2 white pearls. 12/2 = 6
I then just fixed the black pearl drew out the possible outcomes when the blue pearl is next to it and there are 3 possible ways to fix them next to each other.
3/6 = 1/2
Took me too long for a problem which might seem simple at first.
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Paul

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OA=1/2
Took me ages too.
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I'm not sure why you need to divide by two because there are two white pearls, how does this change the fact that there is a total of 5 pearls, does it mater if 1,2 or three are white, all that matters is the there is 1 out of 5 that is black and 1 out of 5 that is blue, right?
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jeremy02 wrote:
I'm not sure why you need to divide by two because there are two white pearls, how does this change the fact that there is a total of 5 pearls, does it mater if 1,2 or three are white, all that matters is the there is 1 out of 5 that is black and 1 out of 5 that is blue, right?

Below you will see a diagram which will explain what I mean. In a ring like formation, there are (n-1)! ways of arranging the components by fixing 1 item down. In the drawing, let's say that the black pearl is anchored, there are 4! ways of arranging the remaining items. However, because there are 2 white pearls, you can "interchange them" and you will have to further reduce by 2! the possible number of outcomes. Now, we are down to 4!/2! = 12.

The next concept is a bit more subtle. Normally, when we are fixing people around a round table, (n-1)! would suffice to give you the answer. However, we are talking about beads on a necklace. Once you have one formation, you can "flip" it over and you will get the "mirror" image. Look carefully at the diagram and you will see that although the two necklaces are arranged differently, the right-hand side one is just the "mirror" image of the left-hand side one. Hence, you have to divide everything by another factor of 2 because of the possibility of making another outcome merely by flipping the necklace over. 12/2 = 6
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Paul

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i did this in a diff way...

first the black ball is picked.... then the porbablity that the blue ball will be picked is 1/4

but there are two ways of doing this so...

I got the answer as 1/4 * 2 =1/2

is this a right way of doin it ?
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Jim

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total possible outcome is (n-1)! / 2 = 4!/2 = 12.

Black and Blue next to each other (BB):

1. WBBR
2. RBBW
3. BBRW
4. BBWR
5. WRBB
6. RWBB

BB could be in the middle; to the left; or to the right (relatively speaking).

Pr = favorable outcome / total outcome = 6/12 = 1/2.
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This one is nice ! From where it is Oxon ?
Manager
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My explanation is:

No of ways to arrange n pearls in a necklace: (n-1)! Because two pearls have same color, so total ways are (n-1)!/2 = (5-1)!/2 = A

If blue and black stands together, so we can consider combination of blue and black pearls as one set. Now we have 4 pearls (3 individual and 1 set). Ways to arrange 4 those pearls are (4-1)!/2 (reasoning is the same as above).

Addtionally, within the set we have two ways to turn 2 pearls around. Hence total of possible ways to arrange those pearls are

2*(4-1)!/2 = B

Probability is B/A = 0.5
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twixt,
Got it from an old A-level text book.
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