The question is:

In a game of poker, five players are each dealt $5$ cards from a $52$-card deck. How many ways are there to deal the cards?

The answer is $\frac{52P5}{(5!)^5}$ or $\frac{52!}{47!5!}$ I understand the reason after seeing the answer: $25$ ordered cards are picked from the deck of $52$, then divided by the number of ways each hand can be ordered, because the order of a hand does not matter.

However, before seeing the answer I came up with this attempt:

You choose $5$ cards from $52$, the order doesn't matter so we have $52\choose{5}$. Then you're left with $47$ cards and for the next hand we have $47\choose{5}$. The final sum is ${{52}\choose{5}} + {{47}\choose{5}} + {{42}\choose{5}} + {{37}\choose{5}} + {{32}\choose{5}}$.

This sum is considerably larger, but while I understand the reason $\frac{52!}{47!5!}$ is the correct, I can't find the mistake in my first attempt.

  • 1
    $\begingroup$ Why are you adding? The universal principle of counting states the count of ways to do a series of tasks is the product of counts for each task. $\endgroup$ Oct 27, 2016 at 23:52

1 Answer 1


There are $52!$ ways to shuffle the deck.   Dealing 5 cards to each player, every hand is one of $5!$ arrangements, and there are $27$ cards remaining, whose order we also don't mind.

Thus there are $\frac{52!}{27!~5!^5}$ ways to deal distinct hands to the five players.

Your attempt was adequate except that you need to multiply rather than add.

$$\binom{52}{5}\binom{47}{5}\binom{42}{5}\binom{37}{5}\binom{32}{5}\binom{27}{0} = \dfrac{52!}{5!^5~27!}$$

PS: $\frac{52!}{47!~5!}$ counts the ways to deal distinct hands to one player; $\frac{47!}{42!~5!}$ counts ways to deal to the next player, et cetera...

  • $\begingroup$ Thank you! It all makes sense now. $\endgroup$
    – Y. Sargis
    Oct 28, 2016 at 0:18

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