Discussion
There are two identical red, two identical black and two identical white balls. In how many different ways can the six balls be placed in a row of six cells, one ball per cell, so that balls of the same colour never occupy two consecutive cells?
Answer: D
Count all arrangements, then remove those with a same-coloured pair adjacent.
\[ \text{total} = \frac{6!}{2!\,2!\,2!} = \frac{720}{8} = 90 \]
Treating one colour's pair as a single block:
\[ \frac{5!}{2!\,2!} = 30 \quad \text{for each of the 3 colours} \]
Two colours blocked:
\[ \frac{4!}{2!} = 12 \quad \text{for each of the 3 pairs of colours} \]
All three blocked: \( 3! = 6 \).
By inclusion–exclusion the number with at least one adjacent pair is
\[ 3(30) - 3(12) + 6 = 90 - 36 + 6 = 60 \]
\[ 90 - 60 = 30 \]
All three correction terms are needed; stopping after the first subtraction would leave zero.
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