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.
I will assume that the counters of the same color cannot be told apart,
i.e., are indistinguishable. If this is a false assumption, and they are
to be considered as distinguishable, then tell me in the thank-you note
form below and I'll get back to you by email. (No charge ever, as I do
this for fun and to help you).
First we place the non-red ones any of these 3C1 = 3 ways:
[(3 places choose 1) ways to place the white one. The
3 ways are
1. right of,
2. between, and
3. left of the two blue ones.
B B W
B W B
W B B
Then for each of these ways of arranging the 3 non-red ones, there
are 4 places to put the 3 red ones:
1. to the left of the left-most non-red one.
2. between the left-most non-red one and the middle non-red one.
3. between the middle non-red one and the right-most non-red one.
4. to the right of the right-most non-red one.
That's (3C1)(4C3) = (3)(4) = 12 ways.
The 12 ways are:
1. RBRBRW
2. RBRBWR
3. RBBRWR
4. BRBRWR
5. RBRWRR
6. RBRWRR
7. RBWRRR
8. BRWRRR
9. RWRBRB
10. RWRBBR
11. RWBRBR
12. WRBRBR
Be sure to let me know if the counters are distinguishable.
Edwin