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Permutation and Combinations

In a row of 6 counters, 3 are red, 2 are blue and 1 is white.
Find the number of arrangements in which no red counter is
next to another red counter.
 
Ans is 12
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2 Answers

Let's say the B and W counters are the separators for the R ones.

You have this _B_B_W_ (where _ is a space) so there are 4 places to put 3 R's, hence 4C3 = 4 places for R.

For each and every one of these selections on where you place the 3 R's, you have 3C1 places to put the W amongst B.
Hence 4x3 = 12 total arrangements.

Enumerating (in each group of 3 I just shuffle the BBW):

R B R B R W    
R B R W R B
R W R B R B

R B R B W R
R B R W B R
R W R B B R

R B B R W R
R B W R B R
R W B R B R

B R B R W R
B R W R B R
W R B R B R





Comments

Good correction.  I missed the cases where a red is at each end.  Daniel, Al has the correct answer.
Daniel,
 
Sure, it comes from this  _B_B_   ( where _ is a space).
 
The W can go in any 1 of the 3 spaces around/between the B's.  Its 3 choose 1.   ( BBW, BWB, and WBB )
 
Hope this helps!
 
 
There are basically two cases.  Letting N represent not red, we can either have:
 
RNRNRN or NRNRNR
 
Therefore, we need to figure out the ways of arranging 2 blue and 1 white into 3 "N" slots.  There are three possibilities for this, corresponding to each of the positions we can place the white.  Therefore, since there are 2 cases, we have 2*3 = 6 possible combinations that meet the criteria.
 
If you have additional questions on combinatorics, please feel free to let me know.

Comments

 
 RNRNNR and RNNRNR  are also valid combinations