deck of cards

my math teacher gave us this problem( among others) to solve for homework, even though she hasn't even started teaching the section on probability and statistics :x ... i would like some help please. and i have NO idea how to do this :? by the way, so explain it if you can.

list all the possible combinations of 2 cards that would add to 6 in a standard 52 card deck.

thanks

Start by doing a bit of logical "thinking":
only 2-card-combos that can add to 6 are: 1,5 and 2,4 and 3,3 ; agree?
Now play with that...

Hello, wortwortwort117!

. . . and i have NO idea how to do this . . Really? That's sad.

List all the possible combinations of two cards that would add to 6 in a standard 52-card deck.

Click to expand...

If you have a deck of cards, you'd quickly learn that there are three basic combinations for a sum of six:
. . Ace and Five, Two and Four, Three and Three.


Consider the sum "Ace and Five".
There are 4 choices for the suit of the Ace and 4 choices for the suit of the Five.

Can you see that there are 16 possible combinations?

. . \(\displaystyle \begin{array}{cccc}A\spadesuit 5\spadesuit & A\heartsuit5\spadesuit & A\clubsuit5\spadesuit & A\diamondsuit5\spadesuit \\ A\spadesuit 5\heartsuit & A\heartsuit 5\heartsuit & A\clubsuit 5\heartsuit & A\diamondsuit 5\heartsuit \\ A\spadesuit 5\clubsuit & A\heartsuit5\clubsuit & A\clubsuit 5\clubsuit & A\diamondsuit 5\clubsuit \\ A\spadesuit 5\diamondsuit & A\heartsuit 5\diamondsuit & A\clubsuit 5\diamondsuit & A\diamondsuit 5\diamondsuit \end{array}\)


And there are 16 combinations for the "Two and Four":

. . \(\displaystyle \begin{array}{cccc} 2\spadesuit 4\spadesuit & 2\heartsuit 4\spadesuit & 2\clubsuit 4\spadesuit & 2\diamondsuit 4\spadesuit \\ 2\spadesuit 4\heartsuit & 2\heartsuit 4\heartsuit & 2 \clubsuit 4\heartsuit & 2\diamondsuit 4\heartsuit \\ 2\spadesuit 4\clubsuit & 2\heartsuit 4\clubsuit & 2\clubsuit 4\clubsuit & 2\diamondsuit 4\clubsuit \\ 2\spadesuit 4\diamondsuit & 2\heartsuit 4\diamondsuit & 2\clubsuit 4\diamondsuit & 2\diamondsuit 4\diamondsuit \end{array}\)


But there are only 6 combination for the "Three and Three":

. . \(\displaystyle \begin{array}{cccc}3\spadesuit3\heartsuit & 3\spadesuit 3\clubsuit & 3\spadesuit 3\diamondsuit \\ & 3\heartsuit 3\clubsuit & 3\heartsuit3\diamondsuit \\ && 3\clubsuit 3\diamondsuit \end{array}\)