SOLUTION: I have 52 cards in a deck. 5 of those cards are in my hand. One's an ace and the rest are not. How many combinations can I have? (How do I solve it?)
Algebra ->
Permutations
-> SOLUTION: I have 52 cards in a deck. 5 of those cards are in my hand. One's an ace and the rest are not. How many combinations can I have? (How do I solve it?)
Log On
Question 597893: I have 52 cards in a deck. 5 of those cards are in my hand. One's an ace and the rest are not. How many combinations can I have? (How do I solve it?) Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! There are 4 aces in the deck and 48 non aces.
Combination=nCr=n!/((n-r)!r!)
4C1*48C4
=4*194580
=778320 combinations
.
Ed
