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# What are the total number of ways, in which 3 red and 3 blue

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What are the total number of ways, in which 3 red and 3 blue [#permalink]

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12 May 2004, 18:53
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What are the total number of ways, in which 3 red and 3 blue beads can be put in a circular bracelet?

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12 May 2004, 20:58
3 ways assuming all red beads are identical and all blue beads are identical

Following ways are possible
BBBRRR - You can rotate them to get different combinations
BBRBRR
BRBRBR
These 3 basics arrangements cover all possible arrangements

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12 May 2004, 21:29
anandnk wrote:
3 ways assuming all red beads are identical and all blue beads are identical

Following ways are possible
BBBRRR - You can rotate them to get different combinations
BBRBRR
BRBRBR
These 3 basics arrangements cover all possible arrangements

I think believe my answer is wrong and 4 should be it by laying out the possible arrangements.
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Paul

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12 May 2004, 22:23
Yeah I agree I missed RBBRRB or BBRRBR

I dont know how to solve this problem using formulas.

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12 May 2004, 22:44
anandnk wrote:
Yeah I agree I missed RBBRRB or BBRRBR

I dont know how to solve this problem using formulas.

Anand, as a programmer, you could write some algorithm to solve it. My friend solved this problem for N red and N green (it was given at all-Kazakh programming contest ). Kazakhstan is a large country in Central Asia (former Soviet Union).

4.

The method is as follows:

1. Think about combinations where all Bs are actually non-sequential. There is only 1 such comb.

2. Think about combs where only 2 Bs are sequential, but the remaining is not. => 2 combs.

3. Think about combs where all three B are sequential => there is only 1 such comb.

The general method for (N greens, N reds) is as follows:

1. with no neighbors = 1.

2. with only 1 pair of 2 neighbors = N-1.

3. with only 2 pairs of 2 neighbors, ... etc.

So, it can be easily done even for N = 4,5.

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13 May 2004, 10:35
I am not getting this . any help will be appreciated.
I got 8.
Any circular bracelet with six positions can be imagined as

_ _ _ _ _ _

we are required to sit 3 R balls and 3 B balls in six spaces.

so 2 X 2 X 2 X 1 X 1 X 1 = 8
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Chetan

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13 May 2004, 10:49
sunny_god76 wrote:
I am not getting this . any help will be appreciated.
I got 8.
Any circular bracelet with six positions can be imagined as

_ _ _ _ _ _

we are required to sit 3 R balls and 3 B balls in six spaces.

so 2 X 2 X 2 X 1 X 1 X 1 = 8

You got 8 because symmetric allocations were counted TWICE.

RBRBRB == BRBRBR! You did the job twice...

A propos, how did you get 2 X 2 X 2 X 1 X 1 X 1? It should rather be C[3,6] = 6!/(3!*3!) = 24... The fact that there are problems with symmetry makes this problem not so easy. As I've already said, my friend wrote a program to calculate total # of combinations in general case...

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13 May 2004, 10:49
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