<s>[INST] Probability of selecting a poker hand I am trying to solve a probability problem about five-card poker hand. I have access to the answer which is different from what I had come up with. The question is: What is the probability that a five-card poker hand has exactly two cards of same value, but no other cards duplicated?
My answer to this question was as follows:
$\binom{13}{1} \binom{4}{2} \binom{48}{1}\binom{44}{1} \binom{40}{1}$.
Which means:

*

*First select a card number then select its two suits ie. $\binom{13}{1} \binom{4}{2}$. These will be the two cards of same value.

*Select three other cards which are not duplicate as: $\binom{48}{1}\binom{44}{1} \binom{40}{1}$.

The correct answer doesn't match my answer. This answer is provided in book AOPS and is as: $\binom{13}{1} \binom{4}{2}\binom{12}{3}\binom{4}{1}\binom{4}{1}\binom{4}{1}$.
So question is, what am I doing wrong? Thanks
 [/INST] You and the book count differently how to select the three remaining cards. Your answer is:
$$ \binom{48}{1}\binom{44}{1} \binom{40}{1} = 48 \cdot 44 \cdot 40 = 4^3 \cdot 12 \cdot 11 \cdot 10$$
The book answer is:
$$\binom{12}{3}\binom{4}{1}\binom{4}{1}\binom{4}{1} = 4^3 \cdot \frac{12\cdot 11\cdot 10}{3!}$$
They differ by a $3!$ factor, which is precisely the number of permutations of three distinct objects. This suggests that you are considering the order of the three remaining cards.
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