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As a programmer, discovering the binomial theorem has helped me with alot. I want to solve something using maths as source and I wonder if you can help me define this one:

Regarding this question and its following outcome with: $${20 \choose 10} = 184756$$

I want to add further combinations. Say, every blue ball has a random letter written on it, A-Z(26 combinations), and red balls a number 0-9(10 combinations).

The letters and numbers cannot be repeated in once instance. e.g. AABCDEFGHI1234567890 is not allowed because 'A' is repeated.

There can be only ten blue and ten red balls in the bag.

How do I express this mathematically to calculate all the possible combinations?

Community
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OHMR
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  • thank you, I made the edit – OHMR Sep 10 '15 at 21:23
  • Do you still have ten balls of each colour? How are the labels distributed? Perhaps it would be best to simply re-ask the question with the new parameters. – 727 Sep 10 '15 at 21:46
  • yes you are right, im missing out on some params. edited again, sry – OHMR Sep 10 '15 at 23:11

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So unless I'm misunderstanding something, you're essentially asking for the number of possible strings of length $20$ where exactly $10$ of the characters are digits and the other $10$ are letters, with no repetitions.

I think we can simply modify the previous solution. There are $\binom{20}{10}$ ways to place the red and blue balls (digits and letters). In each configuration, the first blue ball to be labelled has $26$ possible labels, the second has $25$, and so on. The first red ball to be labelled has $10$ possible labels, the second has $9$, and so on. In total this gives us $\binom{20}{10} \cdot \frac{26!}{(26-10)!} \cdot \frac{10!}{(10-10)!}$ possible strings.

727
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  • Oops, I just saw that we are not allowed repetition. Let me edit quickly. – 727 Sep 10 '15 at 23:38
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    This looks almost correct. I think you should erase the $+1$ in both denominators, since essentially you are computing permutations as in $$26P10=\frac{26!}{(26-10)!}=\frac{26\cdot25\cdot...\cdot2\cdot1}{16\cdot15\cdot...\cdot2\cdot1}=26\cdot25\cdot...\cdot18\cdot17$$ and similarly for $10P10=10!$. Note that $(10-10)!=0!=1$ by definition. The definition is made so that $0!$ becomes neutral wrt. multiplication thus being the neutral element $1$ in that respect. – String Sep 11 '15 at 09:21
  • @String Right, I miscounted. Thank you. – 727 Sep 11 '15 at 22:34