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Please, check whether I'm correct. Consider a standard deck of cards.

Firstly, how many ways can we equally deal whole deck among $4$ people? Fairly simple question, for we choose $13$ cards from the whole deck, $4$ times. Then we divide the answer by $4!$, for each ordering, which should finally equal to $$\binom{52}{13}*\binom{39}{13}*\binom{26}{13}*\binom{13}{13}\over4!$$

Secondly, how many ways can we pick $13$ cards, such that the choice includes at least one of each suit. That should be $$\binom{4}{1}*\binom{13}{1}*\binom{13}{1}*\binom{13}{1}*\binom{48}{9}$$

since there are four suits, that gives us 1 card to choose from $4$ suits, or $\binom{4}{1}$ ways. Then, for each of the other three suits we can choose in $\binom{13}{1}$ ways. Finally that leaves us with $48$ other cards and $9$ other draws.

Last question is a combination of the previous two. How many ways can we deal cards like in first example, with only exception being that the first player has to have at least one of each suit, like in second example. Using the same logic, that should result in $$\binom{4}{1}*\binom{13}{1}*\binom{13}{1}*\binom{13}{1}*\binom{48}{9}*\frac{\binom{39}{13}*\binom{26}{13}*\binom{13}{13}}{3!}$$

Are my findings correct? Please, point out any mistakes you saw in this. Thank you in advance.

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Your count of $13$ cards with at least one of each suit is not correct. You are multiple counting every hand. If a hand has the ace, 2, and 3 of spades, you count it once when the ace is the first spade and the 2 and 3 are in the remaining nine cards, then again when the 2 is the first spade and again when the 3 is the first spade.

Once you fix the problem with the second, the third becomes correct if you make the denominator $3!$ because there are only three other hands you can permute.

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    $\begingroup$ I don't agree with permuting the hands. In a game of bridge, for example it's very important who gets which hand. Even a cyclic permutation is inadmissible in that case, since the dealer bids first, so it's important which hand he gets. $\endgroup$ – saulspatz Dec 1 at 16:30
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    $\begingroup$ @saulspatz: OP was allowing permutation, so I went with that. I agree that most games do not allow permutation, but this is just dividing the cards into indistinguishable piles. $\endgroup$ – Ross Millikan Dec 1 at 17:11
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In how many ways can a standard deck be dealt to four people so that each person receives $13$ cards?

Since it matters which person receives which cards, the cards may be distributed in $$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$$ ways. It does not make sense to divide by $4!$ unless it does not matter who receives which cards.

In how many ways can $13$ cards be selected so that the choice includes at least one card of each suit?

There are $$\binom{52}{13}$$ ways to select $13$ cards from the deck. From these, we must subtract those selections in which one or more suits is missing.

A suit is missing: There are $\binom{4}{1}$ ways to exclude one of the four suits and $\binom{39}{13}$ ways to select $13$ cards from the remaining suits. Hence, there are $$\binom{4}{1}\binom{39}{13}$$ ways to select $13$ cards so that one of the suits is excluded.

However, if we subtract this amount from the total, we will have subtracted each selection in which two suits are missing twice, once for each way we could have designated one of the two missing suits as the missing suit. We only want to subtract these cases once, so we must add them back.

Two suits are missing: There are $\binom{4}{2}$ ways to exclude two of the four suits and $\binom{26}{13}$ ways to select $13$ cards from the remaining two suits. Hence, there are $$\binom{4}{2}\binom{26}{13}$$ ways to select $13$ cards from the deck so that two suits are excluded.

If we subtract those hands from which a suit is missing and then add those hands from which two suits are missing, we will not have subtracted those cases in which three suits are missing. This is because we subtracted them three times, once for each way we could have designated one of the three missing suits as the missing suit, and then added them three times, once for each of the $\binom{3}{2}$ ways we could have designated two of the three missing suits as the two missing suits. Thus, we must still subtract those hands in which three suits are missing.

Three suits are missing: There are $\binom{4}{3}$ ways to exclude three of the suits and one way to take all $13$ cards of the remaining suit. Hence, there are $$\binom{4}{3}\binom{13}{13}$$ ways to select $13$ cards from the deck if three suits are excluded.

It is not possible to exclude all four suits and still draw $13$ cards.

By the Inclusion-Exclusion Principle, the number of ways $13$ cards may be selected from the deck so that there is at least one card from each suit is $$\binom{52}{13} - \binom{4}{1}\binom{39}{13} + \binom{4}{2}\binom{26}{13} - \binom{4}{3}\binom{13}{13}$$

In how many ways a standard deck be dealt to four people so that each person receives $13$ cards and the first person receives at least one card of each suit?

Multiply the number of ways of distributing $13$ cards to the first person so that he or she receives at least one card of each suit by the number of ways of distributing the remaining $39$ cards in the deck so that each person receives $13$ of those cards.

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