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I was asked for the probability of having at least one card from each type (hearts/spades/etc.) in a 5 cards hand chosen randomly from a 52 cards normal deck. my solution was $\frac{{{13}\choose{1}}^4{{48}\choose{1}}}{{52}\choose{5}}$ as in, choose one card from each type and then one from the 48 cards left, then divide it all by all the possible hands. my solution is wrong, and I don't understand why, can someone please point out where have I gone wrong?

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marked as duplicate by José Carlos Santos, Claude Leibovici, Arnaud D., Namaste, Krish Nov 21 '17 at 14:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Welcome to MathStackExchange. Thanks for using MathJax it will help the rest of us better able to read your work. The standard word that we use in the US for what you call "type" is "suit". So if you already searched for a similar question you may want to try again using suit. $\endgroup$ – Stephen Meskin Nov 21 '17 at 6:51
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    $\begingroup$ Thank you for your comments and clarification. I was mostly interested in knowing why my approach failed, the correct answer in the given link (which I indeed haven't found while not knowing the word "suit") is also presented in my book. so I am not sure if to mark this as duplicated. $\endgroup$ – Fuseques Nov 21 '17 at 7:20
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You have double counted every single hand, so your answer is twice as large as it should be.

For example, look at the hand consisting of the four aces and the king of hearts. It is counted once when the four aces are among the first four cards, and the king is the final card. However, it is also counted once when the king and the the other three aces are picked first, and the ace of hearts is the final card.

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If you are keeping the exponent $^4$, all that you need to do to get the correct answer is to divide by $2$

This is because one suit will have $2$ cards, and you are counting both (say) $10-8 \;\; and \;\; 8-10$

A better way, though, is to count as given in the first part of the "duplicate"

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