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I think the valid combinations are, B/B/B, G/G/G, B/G/G and G/B/B. Total valid number of combinations are $4$.

My question is whether I should be using [combination theory with repetition] (https://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition) for this problem, I think it falls into combination with repetition scenario (see above link), where $N=2$ (B and G) and $k = 3$ (select $3$ kids), and $\big(\!\binom{N}{k}\!\big) = 4$, wondering if that is the correct theory to use for this problem? Thanks.

Regards,
Lin

Lin Ma
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1 Answers1

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If the order is not considered there are only 4 combinations, no boys, one boy, two boys, three boys

If the order is considered then there are $ 2^3 $ different combinations

The first child could be boy or girl, so two possibilities Similarly for the next two

The total comes up to $2 \times 2 \times 2 = 2^3 =8$

Karthik Vasu
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  • Thanks Karthik, nice catch and order is not considered. Vote up for your reply and wondering your comments on my original question, i.e. "Using combination theory, I think it falls into combination with repetition scenario, where N=2N=2 (B and G) and k=3k=3 (select 33 kids), and ((Nk))=4((Nk))=4, is that correct calculation using the right theory?" Thanks. – Lin Ma Jun 06 '16 at 02:35
  • Hi Karthik, if could comment on my above question, it will be great. :) – Lin Ma Jun 07 '16 at 18:33
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    Can you explain what your question is again – Karthik Vasu Jun 07 '16 at 19:03
  • Sure Karthik, I edit my original question for clarifying what is my original question -- i.e. whether using the right combination theory for the calculation (not asking for the combination result, but whether using right theory). Your advice is appreciated and vote up. :) – Lin Ma Jun 08 '16 at 06:14
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    Yes it is the correct use of the theory. – Karthik Vasu Jun 08 '16 at 06:46
  • Thanks Karthik for the confirm, vote up and mark your reply as answer. – Lin Ma Jun 08 '16 at 18:16