Permutation and Combinations

Question

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

Answer 1

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

Answer 2

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.

Answer 3

Good work, Al P. I now see that the correct answer is 12. This question was posted 2 times by the same person, which makes twice as much work keeping up with answers. Essentially, I missed the two additional combinations that you cite. Students are requested to only submit a specific question ONE TIME!