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I was looking at this question Probability of selecting four letters from ENCYCLOPAEDIA and my mind started to wander.

What would you have to do to handle the double B in question 2 compared to question 1.

  1. You have a bag of 5 letters. A,A,B,C,D. If you choose 3 How many combinations are there with exactly one A.

  2. You have a bag of 5 letters. A,A,B,B,C. If you choose 3 How many combinations are there with exactly one A

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  • $\begingroup$ @Rafael Thank you. $\endgroup$ – Tolure Feb 21 at 16:35
  • $\begingroup$ The answer to question $1$ is $3$ and the answer to question $2$ is $2$. I'm not sure what you're driving at. $\endgroup$ – saulspatz Feb 21 at 16:41
  • $\begingroup$ Looking for how to approach the question (work involved) not just the answer which I could do by writing them all out. $\endgroup$ – Tolure Feb 21 at 16:48
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You approach it like any combinatorics problem, you eliminate possibilities. If only 1 A is required, ignore the rest. Then work out how many combinations there are for the remaining. In the first case that's ABC ABD ACD. In the second we can't ignore the second B, so we get ABB and ABC. There would be 6=3*2*1 times as many, if order mattered.

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In first case, in your combination there should be one $A$ and an arbitrary combination of two letters from the three different non-$A$s. You can select them in $3$ different ways. So the answer is $3$.

In the second case you also select one $A$, but then you have only two choices: to select a $B$ and a $C$, or to select two $B$-s. So the answer is $2$ now.

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  • $\begingroup$ I was looking for a more general application. The question uses a simple example just to limit the noise. Lets say you had m As and n Bs and x Cs, and y ... how would you approach it. $\endgroup$ – Tolure Feb 21 at 16:46

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