SOLUTION: What is the probability that a poker hand (5 card draw) contains nothing but aces and eights? -I've went to my instructor for help and he told me that it would be a combination

Question 725503: What is the probability that a poker hand (5 card draw) contains nothing but aces and eights?
-I've went to my instructor for help and he told me that it would be a combination problem.
-The way I was doing it was that I wrote out the elements:
{1(A),4(8)},{2(A),3(8)},{3(A),2(8)},{4(A),1(8)}
Then I wrote my answer as (1/52)*(1/51)*(1/50)*(1/49)*(1/48) because a poker hand is without replacement (showing that I am not returning an ace or eight back to the deck).
-I've tried both combination formulas that I have and it is not giving me an answer less than or equal to 1 (since it is a probability problem).
the two combination formulas I have are:
n!/[r!(n-r)!] (without replacement)
(r+(n-1))
( r ) (with replacement) *(trying to show a combination "()" instead of a fraction or two separate "()")*

Found 2 solutions by lynnlo, stanbon:
Answer by lynnlo(4176) (Show Source): You can put this solution on YOUR website!

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
What is the probability that a poker hand (5 card draw) contains nothing but aces and eights?
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I'm assuming there are at least one ace and 1 eight in each hand.
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Successful combinations in ace/8's order:
1/4 ; 2/3 ; 3/2 ; 4/1
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# of ways to get each combination:
1/4::: 4C1*4C4 = 4*1 = 4
2/3::: 4C2*4C3 = 7*4 = 28
3/2::: 4C3*4C2 = 4*7 = 28
4/1::: 4C4*4C1 = 1*4 = 4
Total number of 5 card hands with at least one ace and one eight: 64
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Total number of 5 card hands: 52C5 = 2598960
Probability = 66/2598960 = 0.00002539
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Cheers,
Stan H.
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