This question pertains to a standard deck of playing cards. If you are unfamiliar with such cards, refer to the "Probability of an Event" section, specifically the "Standard Playing Cards" heading. A five-card hand is a selection of 5 cards from the 52-card deck. 1. How many different five-card hands can be drawn from a standard deck of 52 playing cards? 2. How many five-card hands contain exactly two hearts? 3. How many five-card hands consist solely of hearts and diamonds? 4. How many five-card hands have four cards of the same rank? 5. A "full house" is a five-card hand with two cards of one rank and three cards of another rank, such as a hand comprising the queen of hearts, queen of spades, 8 of diamonds, 8 of spades, and 8 of clubs. How many five-card hands contain a full house? 6. How many five-card hands do not include any pairs (i.e., no two cards of the same rank)?

1. To determine the number of different five-card hands that can be drawn from a standard deck of 52 playing cards, we use combinations because the order of the cards does not matter. The formula for combinations is given by C(n, k) = n! / (k!(n-k)!), where "n" is the total number of items to choose from, "k" is the number of items to choose, and "!" denotes factorial. For a five-card hand from 52 cards, n = 52 and k = 5. C(52, 5) = 52! / (5! * 47!) = 2,598,960 different five-card hands.

2. To calculate the number of five-card hands that contain exactly two hearts, we choose 2 hearts from the 13 available and then choose 3 more cards from the remaining 39 non-heart cards. C(13, 2) * C(39, 3) = (13! / (2! * 11!)) * (39! / (3! * 36!)) = 78 * 9,139 = 712,842 hands.

3. To calculate the number of five-card hands that consist solely of hearts and diamonds, we need to select all 5 cards from the 26 available hearts and diamonds. C(26, 5) = 26! / (5! * 21!) = 65,780 hands.

4. For a hand with four cards of the same rank, we first select the rank (there are 13 choices) and then pick 4 of the available suits. After that, we need to choose one more card from the remaining 48 cards. 13 * C(4, 4) * C(48, 1) = 13 * 1 * 48 = 624 hands.

5. A full house contains 3 cards of one rank and 2 cards of another rank. There are C(13, 1) ways to choose the rank of the three-of-a-kind, C(4, 3) ways to choose the suits for that rank, C(12, 1) ways to choose the rank of the pair, and C(4, 2) ways to choose the suits for the pair. C(13, 1) * C(4, 3) * C(12, 1) * C(4, 2) = 13 * 4 * 12 * 6 = 3,744 hands.

6. To have a hand with no pairs, we must choose 5 different ranks and then one suit for each of these ranks. There are C(13, 5) ways to choose the ranks. For each rank, there are C(4, 1) ways to choose a suit. Since the choices are made independently for each rank, we multiply the number of ways to choose the suits. C(13, 5) * (C(4, 1))^5 = (13! / (5! * 8!)) * (4^5) = 1,287 * 1,024 = 1,317,888 hands.