Find the number of arrangements in which no red counter is

next to another red counter.

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.

Find the number of arrangements in which no red counter is

next to another red counter.

Ans is 12

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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):

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 =

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

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

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.

RNRNNR and RNNRNR are also valid combinations

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