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If from a deck of 52 cards, I extract 10. In how many combinations do you get at least one ace?

I have come up with two possible answers, but I don't know which one is the right one and why.

So one is $$4{51 \choose 9}$$ the reasoning is that first I extract an ace from the four that there are, and then I have ${51 \choose 9}$ combinations for the other 9 cards.

The second is $${52 \choose 10} - {48 \choose 10}$$ reasoning that there are ${52 \choose 10}$ total combinations of ten cards and I subtract ${48 \choose 10}$ combinations without any aces.

So from other questions it seems that the first one is the correct answer, but why would the second one be wrong?

Thanks for the answers.

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    $\begingroup$ Your first method counts hands with more than one ace multiple times. $\endgroup$ – lulu Nov 27 '18 at 16:49
  • $\begingroup$ @lulu That's important as a point, but the question wants to count all hands with at least one ace. There is still overcounting going on though. $\endgroup$ – Namaste Nov 27 '18 at 16:50
  • $\begingroup$ @amWhy lulu is pointing out that certain hands are counted too many times. $\endgroup$ – Théophile Nov 27 '18 at 16:53
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    $\begingroup$ The second method is correct. It is the best way for the general question. - to obtain probability of at least one, calculate the probability of none and subtract from $1$. $\endgroup$ – herb steinberg Nov 27 '18 at 17:00
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    $\begingroup$ @amWhy Not sure I see your point. if $\{X_i\}_{i=1}^8$ denote specific non-aces, then the OP's method counts $A\spadesuit, A\heartsuit, \{X_i\}_{i=1}^8$ as one case, and the equivalent reordered hand $A\heartsuit, A\spadesuit, \{X_i\}_{i=1}^8$ as another. I understand the OP is counting hands containing at least one ace. $\endgroup$ – lulu Nov 27 '18 at 17:01
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The first approach needs to be modified to ${4\choose1}{48\choose 9}+{4\choose2}{48\choose 8}+{4\choose 3}{48\choose 7}+{4\choose 4}{48\choose 6}$, for instance.

The second is correct.

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To see why your first answer is not correct, assume you have a deck of 3 with 2 aces and you'll draw 2 cards with at least one ace. Call the cards $A_1,A_2,B$

Based on your logic, you compute $2 {2 \choose 1}=4$ combinations. However, clearly you have only $3$. This is due to double counting $\{A_1,A_2\}$

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