SOLUTION: Among the 2,598,960 possible five-card poker hands from a standard 52-card deck, how many contain AT LEAST ONE CLUB (complement of "no clubs")?

Question 649593: Among the 2,598,960 possible five-card poker hands from a standard 52-card deck, how many contain AT LEAST ONE CLUB (complement of "no clubs")?
Found 2 solutions by ewatrrr, swincher4391:
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

Hi,
five-card poker hands from a standard 52-card deck, how many contain AT LEAST ONE CLUB (complement of "no clubs")?
52C5 - 39C5 = 2598960 - 575757
Answer by swincher4391(1107) (Show Source): You can put this solution on YOUR website!
N[at least one club] = N[total ways] - N'[at least one club] = N[total ways] - N[no clubs]
There are 39 cards without a club. So there are 39 choose 5 = 575757 ways to not get a club.
There are 2598960 possible hands, leaving us with 2598960-575757 ways.
Which is 2023203 hands.

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