+4 A standard deck of playing cards has 52 cards divided into four suits (clubs, diamonds, hearts, spades). Each suit consists of nine "number cards," each containing a different number from 2 to 10 , and four "face cards" that include a jack, a queen, a king, and an ace. In the game of Cribbage, points are earned if you can combine two cards that sum to 15. Jacks, queens and kings each have a value of 10, aces each have a value of 1, and all number cards have a value of the number shown. How many different unordered pairs of two cards sum to in a standard 52-card deck?
+887 We can solve this problem by considering different cases based on the value of one of the cards in the pair:
Card 1 value is 1 (Ace): In this case, the other card must have a value of 14. There's only one possibility: a King of any suit (out of 4 kings). So, there are 1 * 4 = 4 such pairs.
Card 1 value is 2 to 8: The other card's value must be 13 down to (15 - card 1 value). For example, if the first card is a 3, the other card can be a 12 (Queen), 11 (Jack), 10 (King), or 9. There are 4 cards (one from each suit) that fulfill this condition for each value from 2 to 8. So, there are 7 values (from 2 to 8) * 4 cards/value = 28 such pairs.
Card 1 value is 9: The other card's value must be 6. There are 4 cards (one from each suit) with a value of 6. So, there are 1 * 4 = 4 such pairs.
Card 1 value is 10: The other card's value must be 5. There are 4 cards (one from each suit) with a value of 5. So, there are 1 * 4 = 4 such pairs.
Card 1 value is a Face Card (Jack, Queen, or King): In this case, the other card must be a 5 or a 6. There are a total of 4 Jacks, 4 Queens, and 4 Kings (12 cards). Each can be paired with a 5 (4 cards) or a 6 (4 cards) for a total of 2 * 4 = 8 such pairs.
Adding the number of pairs from all the cases:
Total Pairs = Cases 1 + 2 + 3 + 4 + 5 = 4 + 28 + 4 + 4 + 8 = 48
Therefore, there are 48 different unordered pairs of two cards that sum to 15 in a standard 52-card deck.
