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If it is assumed that all (52C5) poker hands are equally likely, what is the probability of being dealt two pairs? (This occurs when the cards have denominations a, a, b, b, c, where a, b, and c are all distinct.

Here I am not getting why can't I solve the question as

(13C1)(4C2)(12C1)(4C2)(11C1)(4C1)

the actual answer is (13C2)(4C2)(4C2)(11C1)(4C1)

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  • $\begingroup$ Welcome to Maths SE. Please use MathJax to format mathematics on this site. $\endgroup$
    – Toby Mak
    Commented Mar 13, 2022 at 6:50
  • $\begingroup$ Also, see this exact same question. It would be useful next time if you could tell us where the question comes from: how to ask a good question. $\endgroup$
    – Toby Mak
    Commented Mar 13, 2022 at 6:55

1 Answer 1

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This is a very common mistake you are committing that has to be guarded against.

When you write $(13C1)(12C1)$ for choosing $a's$ and $b's$, you double count ($ab$ and $ba$)

We need to directly choose a pair , $13C2$ since we are not concerned with the order in which they were chosen

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