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So correct answer is$$ {13\choose1} {4\choose3}{12\choose2}{4 \choose1}{4 \choose1} $$which has pretty simple explanation choose 1 card of a value and choose 3 suits. choose 2 cards of different value from rest and suit can be anything so choose any one suit

Now I am just thinking why we can't do it like this choose 3 cards of different values. and then choose 3 suits for first card and then choose one - one suit for another 2 card value. so, $$ {13\choose3}{4\choose3}{4\choose1} {4\choose1}$$

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You're forgetting about the choice of which of the three values you pick is actually the card that has three copies in the hand. Once we multiply your method by $\binom{3}{1}$, we do get the same answer - $54912$ in both cases.

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  • $\begingroup$ hmm, then why don't we multiply for other value cards (3choose1) for 3 copies card, (2choose1) for other 1 copy card and (1choose1) for last one $\endgroup$
    – Priyanshu Dwivedi
    Commented Oct 25, 2019 at 10:14
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    $\begingroup$ Because picking the value which shows up three times completely determines the hand - once you do that, you know that the other two values both show up on exactly one card and you don't need to do anything else with them. $\endgroup$
    – KReiser
    Commented Oct 25, 2019 at 17:51

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