Hello All,

Please help me solve this problem:

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

- Oct 23rd 2009, 08:10 AMashwinkumardpChoosing 4 cards of different suits and different values from a deck of 52 cards.
Hello All,

Please help me solve this problem:

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 - Oct 23rd 2009, 09:08 AMSoroban
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?

. . $\displaystyle \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 $\displaystyle A\clubsuit$

Then these choices remain for the second card:

. . $\displaystyle \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 $\displaystyle 2\heartsuit$

Then these choices remain for the third card:

. . $\displaystyle \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 $\displaystyle 3\spadesuit$

Then these choices remain for the fourth card:

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

- Oct 2nd 2010, 03:09 PMRelmiw
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?