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Currently while solving exercise question I came to this one which is stated as follows:

A bag contains $5$ black and $6$ red balls. Determine the number of ways in which $2$ black and $3$ red balls can be selected from the lot.

Now my logic while solving this was that since the balls of the same color are 'indistinguishable' therefore the selection of any pair would be indistinguishable leading to just $1$ distinct selection. Whereas the answer is equal to $${5\choose2} \times{6\choose3}= 200$$ which would be obvious if the balls were somehow distinguishable.

So where am I going wrong?

  • You aren't wrong mathematically. If we only care about the colors of the balls, then you're right. If we care about which physical balls are selected, then the book is right. It's a matter of interpretation. (I would interpret the question as the book did, since it's in the chapter on permutations and combinations, right?) – saulspatz Oct 23 '19 at 14:32
  • Yes it is in that chapter. –  Oct 23 '19 at 14:33
  • It doesn't matter if they are indistinguishable, the value calculated is necessary in determining the probability of that selection occurring out of all unique selections. – Legorhin Oct 23 '19 at 14:39

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What you thought is completely correct. The source problem does not clearly state that the balls are, apart from their colors, distinguishable. This means that the black balls are numbered from $1$ to $5$, and the red balls from $1$ to $6$. Given these numberings there are the $200$ listed selections.

Christian Blatter
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