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Suppose we have 3 black balls, 4 white balls, 5 red balls and 6 blue balls and we want to arrange them randomly in a line. I want to find out the probability of choosing an arrangement where all identical coloured balls are adjacent together, with the assumption that balls of the same colour are indistinguishable. I also want to find the same probability with the assumption that balls of the same colour are distinguishable.

My solution:

For the assumption that balls of the same colour are indistinguishable:

We have $\frac{18!}{3!4!5!6!}$ distinct arrangements. Now, since balls of the same colour are indistinguishable and there are $4$ colours of balls then there are $4!$ ways of choosing an arrangement where balls of the same colour are adjacent. The probability of choosing an arrangement where identical coloured balls are adjacent together is thus $\frac{4!^23!5!6!}{18!} = \frac{1}{21441420}$.

For the assumption that balls of the same are distinguishable:

We have $18!$ arrangements. Now, since balls of the same colour are distinguishable and there are $4$ colours of balls then there are $4!(4!3!5!6!)=4!^23!5!6!$ ways of choosing an arrangement where balls of the same colour are adjacent. The probability of choosing an arrangement where identical coloured balls are adjacent together is thus $\frac{4!^23!5!6!}{18!} = \frac{1}{21441420}$.

Now since my answers are equal I realise that I've obviously made a mistake somewhere in my reasoning but I can't exactly where.

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    $\begingroup$ "I want to find out the probability of choosing an arrangement where identical coloured balls are adjacent together." Do you want an arrangement where some pair of two balls of the same color are adjacent? Do you want an arrangment where exactly two balls of the same color are adjacent? Do you want to count arrangements in which all balls of the same color are adjacent? You should be more specific $\endgroup$ – Zubin Mukerjee Jan 14 at 20:09
  • $\begingroup$ @ZubinMukerjee I'll update it, but I thought it was clear from the context. It's where all balls of the same colour are adjacent. $\endgroup$ – Hai Jan 14 at 20:28
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Both your calculations are correct. When you switch from the balls being indistinguishable to distinct, you increase the total number of arrangements and the number of admissible arrangements by the same factor. The proportion of admissible arrangements stays the same. Suppose only the black balls are distinguishable. Then we immediately have $6$ times as many arrangements of all the balls, but we also have $6$ times as many arrangements with balls of identical colors adjacent. The proportion stays them.

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