4. Determine cuántas manos de tercias puede haber en el juego de póker con baraja inglesa (52 cartas, 4 palos y números desde el 1 al 10 , para después pasar a J, Q, K y A). Después, determine cuántas combinaciones distintas de tercias posibles existen sin tomar en cuenta manos de cartas con: a) Números pares b) Números que empiecen con 1 (o sean el número 1) c) Números primos

Alright, I need to determine the number of three-card combinations (tercias) in a standard 52-card English deck. The deck has 4 suits and numbers from 1 to 10, with the numbers 11, 12, and 13 represented by J, Q, and K respectively, and the number 14 represented by A. After calculating the total number of tercias, I also need to find out how many distinct combinations exist without considering hands that have: a) Even numbers b) Numbers that start with 1 (or are the number 1) c) Prime numbers First, let's understand the composition of the deck. There are 4 suits: hearts, diamonds, clubs, and spades. Each suit has 13 cards: numbers 1 through 10, and the face cards J, Q, K, and A. So, total cards = 4 suits * 13 cards = 52 cards. Now, a tercia is a combination of 3 cards. The total number of possible tercias in a 52-card deck can be calculated using combinations since the order of the cards doesn't matter. The formula for combinations is C(n, k) = n! / (k!(n - k)!), where n is the total number of items, and k is the number of items to choose. So, total tercias = C(52, 3) = 52! / (3! * (52 - 3)!) = (52 * 51 * 50) / (3 * 2 * 1) = 22,100. Now, for part a) even numbers: The even numbers in the deck are 2, 4, 6, 8, and 10. There are 5 even numbers per suit, so total even-numbered cards = 4 suits * 5 = 20. The number of tercias with only even numbers = C(20, 3) = 20! / (3! * 17!) = (20 * 19 * 18) / (3 * 2 * 1) = 1,140. For part b) numbers that start with 1 or are the number 1: The numbers that start with 1 are 1, 11, and 12, which are represented by 1, J, and Q respectively. So, total such cards = 4 suits * 3 = 12. The number of tercias with only these numbers = C(12, 3) = 12! / (3! * 9!) = (12 * 11 * 10) / (3 * 2 * 1) = 220. For part c) prime numbers: The prime numbers between 1 and 14 are 2, 3, 5, 7, 11, and 13. So, total prime-numbered cards = 4 suits * 6 = 24. The number of tercias with only prime numbers = C(24, 3) = 24! / (3! * 21!) = (24 * 23 * 22) / (3 * 2 * 1) = 2,024. But wait, I need to ensure that these calculations are accurate and consider whether the restrictions are mutually exclusive or if there's overlap. For example, some numbers might be both even and start with 1, but in this case, the only number that starts with 1 is 1, which is odd, so there's no overlap. Similarly, prime numbers include 2, 3, 5, 7, 11, and 13, none of which are even except for 2, but since we're excluding even numbers in part a), 2 is excluded. Therefore, the calculations for parts a, b, and c) are independent. However, to be thorough, I should consider if there's any double-counting or exclusion errors. Since the restrictions are based on different criteria (even numbers, numbers starting with 1, and prime numbers), and there's no overlap between them, the total number of tercias without these restrictions should be the total tercias minus the tercias that include at least one of these restricted numbers. But the question asks for the number of distinct combinations without considering hands that have these restricted numbers, which means we're looking for the total tercias minus the tercias that include any of the restricted numbers. So, total tercias without restrictions = 22,100. Total tercias with even numbers = 1,140. Total tercias with numbers starting with 1 = 220. Total tercias with prime numbers = 2,024. But since these are mutually exclusive, the total tercias with any of these restrictions = 1,140 + 220 + 2,024 = 3,384. Therefore, tercias without any of these restrictions = 22,100 - 3,384 = 18,716. But wait, this seems too high. Let me double-check the calculations. Total tercias = C(52, 3) = 22,100. Tercias with even numbers = C(20, 3) = 1,140. Tercias with numbers starting with 1 = C(12, 3) = 220. Tercias with prime numbers = C(24, 3) = 2,024. Total tercias with any restriction = 1,140 + 220 + 2,024 = 3,384. So, tercias without any restrictions = 22,100 - 3,384 = 18,716. This seems correct. Therefore, there are 18,716 distinct tercias that do not include any even numbers, numbers starting with 1, or prime numbers.