How many different 4 card hands can be drawn from a set of 30 cards

Assume for now that the ordering of the four cards does matter. Hands like [tex]\verb!A!\, \verb!B!\, \verb!C!\, \verb!D![/tex] and [tex]\verb!A!\, \verb!B!\, \verb!D!\, \verb!C![/tex] would then be considered different from one another.

There would [tex]30[/tex] choices for the first card. Since the first card was not returned to the pile, there would be only [tex]29[/tex] choices for the second card. Likewise, there would be [tex]28[/tex] choices for the third card and [tex]27[/tex] for the fourth.

By this reasoning, there would be [tex]30 \times 29 \times 28 \times 27 = 657,\!720[/tex] different ways to draw a hand of four cards from this set when the ordering of these four cards do matter.

However, in many card games, once a hand of cards is drawn, the ordering of cards within that hand does not matter. In other words, hands like [tex]\verb!A!\, \verb!B!\, \verb!C!\, \verb!D![/tex] and [tex]\verb!A!\, \verb!B!\, \verb!D!\, \verb!C![/tex] would not be considered as distinct from one another.

In that case, the [tex]30 \times 29 \times 28 \times 27 = 657,\!720[/tex] ways of drawing cards would include a large number of duplicates.

There are be [tex]4 \times 3 \times 2 \times 1 = 24[/tex] ways to arrange a hand of four cards when the order matter. Hence, when the ordering within a hand no longer matters, each hand of four cards would have been counted [tex]24[/tex] times among those [tex]30 \times 29 \times 28 \times 27 = 657,\!720[/tex] ways.

Therefore, when the ordering of cards within a set does not matter, [tex]\displaystyle \frac{30 \times 29 \times 28 \times 27}{4 \times 3 \times 2 \times 1} = 27,\!405[/tex] would give the number of distinct ways to draw a hand of four out of this set of thirty distinct cards.