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I was attempting to calculate the number of three of a kind in a five card poker hand. I looked at all the posts pertaining to this question, but none of them had the problem that I currently have in calculating the number of five card poker hands. I reasoned as follows

We want three cards of the same type, so we get

$$\binom{13}{1}\binom{4}{3}\tag{1}$$

Afterwards, I then extract one card from the remaining $12$ types

$$\binom{12}{1} \binom{4}{1}\tag{2}$$

Finally, I then extract one more card from the remaining $11$ types

$$\binom{11}{1} \binom{4}{1} \tag{3}$$

Multiplying $(1),(2)$ and $(3)$ renders $109,824$, which is wrong, and I know and understand the correct answer to be

$$\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2 = 54,912$$

Please offer me insight into what is going on in my answer, and I would greatly appreciate it.

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    Because the number of ways of choosinng 2 out of 12 types is not the same way of choosing one out of twelve types and then one out of eleven types. Your way counts first a king and then a three, as a different thing than first a three and then a king. Your answer needs to consider a king and a three to be one thing; not two. so the answer of choosing two out of twelve things $12 \choose 2$ is a different answer than choosing one thing first and than another ${12 \choose 1} \times {11 \choose 1}$. – 2017-01-22

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Multiplying (1), (2) and (3) renders 27456.

No, multiplying the three values gives 109824, which double counts the correct value.

The reason is that you've counted every hand twice. For example, if we consider a specific poker hand with three aces, a king and a queen, you've counted this once with the king picked before the queen, and again with the queen picked before the king.

Also, you forgot to mention what the original problem was. It is obviously not "calculating the number of five card poker hands" as you say; from your reasoning it seems the problem is calculating the number of five card poker hands where three are the same type and the remaining two are two other types.

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    In your example with counting three aces, a kind, and a queen, and then also counting three aces, a queen, and a king, where in the math is this implied? I'm assuming that the $\binom{12}{1}$ and $\binom{11}{1}$ has something to do with it. – 2017-01-22
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    [The original question has been edited to correct the error you observed.] – 2017-01-22
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    @SteveKass Thanks! – 2017-01-22
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    I see that $\binom {12}{1}\binom {11}{1}$ is equal to the number oof permutations of two objects arranged from twelve objects, but how does this product necessitate an arragement rather than the indistinguishable arrangement/combination that I desire? – 2017-01-24
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    @BenedictVoltaire Not sure what you're trying to say. I already told you the problem with your counting: you counted hands twice. Maybe you'll understand if we have smaller numbers. Consider a deck with three distinct cards. How many two-card hands can be drawn from this deck? It's not $\binom{3}{1}\binom{2}{1}$... – 2017-01-25