Question 469226: Please help me to solve: Suppose that 2 cards are dealt from a standard 52-card poker deck. Let A be the event that the sum of the 2 cards is 8(assume that aces equal 1) How many outcomes are in A?
This is how I started: Combinations that equal 8: A+7, 2+6, 3+5, 4+4, 5+3, 6+2, 7+A. Considering there are 4 cards for each number and add them together is 8 cards per set of combinations. There are 7 combinations, so 7*8=56. How does that sound?
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! There are 4 aces and 4 sevens=4*4=16
Remember that we are looking for combinations not permutations. The order in which
the cards are dealt is not important.
There are 16 possible combinations of A,7.
Another 16 for 6,2 and 16 for 3,5
16*3=48
4C2=6 a special case for 4,4
.
48+6=54
You were close.
.
Ed
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