# Choosing 4 cards of different suits and different values from a deck of 52 cards.

• October 23rd 2009, 08:10 AM
ashwinkumardp
Choosing 4 cards of different suits and different values from a deck of 52 cards.
Hello All,

How many ways are there to choose 4 cards of different suits and different values from a deck of 52 cards?

Thank you,
Ashwin Kumar D P, Bangalore, India
• October 23rd 2009, 09:08 AM
Soroban
Hello, Ashwin!

I had to "draw a picture" to understand this one . . .

Quote:

How many ways are there to choose 4 cards of different suits
and different values from a deck of 52 cards?

There are, of course, 52 cards in the deck.

. . $\begin{array}{cccc}
\clubsuit & \heartsuit & \spadesuit & \diamondsuit \\ \hline
A&A&A&A \\ 2&2&2&2 \\ 3&3&3&3 \\ 4&4&4&4 \\ 5&5&5&5 \\
\vdots & \vdots & \vdots & \vdots \\ K & K & K & K \end{array}$

The first card can be any of the 52 cards.
. . Suppose it is $A\clubsuit$
Then these choices remain for the second card:

. . $\begin{array}{cccc}
\clubsuit & \heartsuit & \spadesuit & \diamondsuit \\ \hline
\boxed{A}&\rlap{X}A&\rlap{X}A&\rlap{X}A \\ \rlap{X}2&2&2&2 \\ \rlap{X}3&3&3&3 \\ \rlap{X}4&4&4&4 \\ \rlap{X}5&5&5&5 \\
\vdots & \vdots & \vdots & \vdots \\ \rlap{X}K & K & K & K \end{array}$

There are 36 choices for the second card.

Suppose the second card is $2\heartsuit$
Then these choices remain for the third card:

. . $\begin{array}{cccc}
\clubsuit & \heartsuit & \spadesuit & \diamondsuit \\ \hline
\boxed{A}&\rlap{X}A&\rlap{X}A&\rlap{X}A \\
\rlap{X}2&\boxed{2}&\rlap{X}2&\rlap{X}2 \\
\rlap{X}3&\rlap{X}3&3&3 \\
\rlap{X}4&\rlap{X}4&4&4 \\
\rlap{X}5&\rlap{X}5&5&5 \\
\vdots & \vdots & \vdots & \vdots \\
\rlap{X}K & \rlap{X}K & K & K \end{array}$

There are 22 choices for the third card.

Suppose the third card is $3\spadesuit$
Then these choices remain for the fourth card:

. . $\begin{array}{cccc}
\clubsuit & \heartsuit & \spadesuit & \diamondsuit \\ \hline
\boxed{A}&\rlap{X}A&\rlap{X}A&\rlap{X}A \\
\rlap{X}2&\boxed{2}&\rlap{X}2&\rlap{X}2 \\
\rlap{X}3&\rlap{X}3&\boxed{3}&\rlap{X}3 \\
\rlap{X}4&\rlap{X}4&\rlap{X}4&4 \\
\rlap{X}5&\rlap{X}5&\rlap{X}5&5 \\
\vdots & \vdots & \vdots & \vdots \\
\rlap{X}K & \rlap{X}K & \rlap{X}K & K \end{array}$

There are 10 choices for the fourth card.

Disgarding the order in which the four cards are drawn,

. . there are: . $\frac{52\cdot36\cdot22\cdot10}{4!} \;=\;\boxed{17,160}$ such 4-card hands.

• October 2nd 2010, 03:09 PM
Relmiw
I understand where the 52, 36, 22, and 10 came from, but why divide by 4!? could anyone describe it in a way that makes sense?