Here is a table showing all 52 cards in a standard deck.Face cardsColor Suit Ace Two Three Four Five Six Seven Eight Nine Ten Jack Queen KingRed Hearts A♥ 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ 9♥ 10♥ J♥ Q♥ K♥Red Diamonds A♦ 2♦ 3♦ 4♦ 5♦ 6♦ 7♦ 8♦ 9♦ 10♦ J♦ Q♦ K♦Black Spades A♠ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠Black Clubs A♣ 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣How many five-card hands (drawn from a standard deck) contain three sixes and two eights?(A five-card hand is any set of five different cards.)(If necessary, consult a list of formulas.)
Question
Here is a table showing all 52 cards in a standard deck.Face cardsColor Suit Ace Two Three Four Five Six Seven Eight Nine Ten Jack Queen KingRed Hearts A♥ 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ 9♥ 10♥ J♥ Q♥ K♥Red Diamonds A♦ 2♦ 3♦ 4♦ 5♦ 6♦ 7♦ 8♦ 9♦ 10♦ J♦ Q♦ K♦Black Spades A♠ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠Black Clubs A♣ 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣How many five-card hands (drawn from a standard deck) contain three sixes and two eights?(A five-card hand is any set of five different cards.)(If necessary, consult a list of formulas.)
Solution
To solve this problem, we need to use the combination formula which is C(n, k) = n! / [k!(n-k)!], where n is the total number of options, k is the number of options chosen, and "!" denotes factorial.
There are 4 sixes and 4 eights in a standard deck of 52 cards.
We want to find the number of ways to choose 3 sixes out of 4 and 2 eights out of 4.
The number of ways to choose 3 sixes out of 4 is C(4, 3) = 4! / [3!(4-3)!] = 4.
The number of ways to choose 2 eights out of 4 is C(4, 2) = 4! / [2!(4-2)!] = 6.
Since these are independent events, we multiply the two results together to get the total number of five-card hands that contain three sixes and two eights.
So, the total number of such hands is 4 * 6 = 24.
Similar Questions
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