💡Alright, so I'm trying to figure out how many different ways the chess match between Westside High School and Eastside High School can be scheduled. Each school has four players, and each player needs to play two games against each player from the opposing school. The match is played over several rounds, with four games happening simultaneously in each round. First, let me break down the problem. Each player from Westside has to play two games against each of the four players from Eastside. So, for one Westside player, that's 4 opponents × 2 games = 8 games. Since there are four players on each team, the total number of games is 4 × 8 = 32 games. Now, since each round consists of four games happening at the same time, the total number of rounds needed is 32 games ÷ 4 games per round = 8 rounds. So, the match will take place over 8 rounds.Next, I need to figure out how many different ways these 32 games can be scheduled into 8 rounds, with each round consisting of four games. This seems like a combinatorial problem where I need to count the number of ways to arrange these games.Let me think about how to model this. Each round consists of four games, each involving one player from Westside and one player from Eastside. Since each player needs to play two games against each opponent, I need to ensure that over the 8 rounds, each pair of players from opposing schools plays exactly two games.One way to approach this is to consider the entire schedule as a sequence of rounds, where each round is a set of four games. The key is to ensure that no two games in a round involve the same player from either school. That is, in each round, each Westside player is paired with exactly one Eastside player, and vice versa.This sounds similar to arranging a round-robin tournament, but with multiple games between each pair. In a standard round-robin tournament, each pair plays once, but here they play twice. So, I need to extend the concept.I recall that in combinatorics, the number of ways to schedule such matches can be calculated using permutations and combinations. Specifically, for each round, we can think of it as a permutation of the Eastside players assigned to the Westside players.Since there are four players on each side, the number of ways to pair them in a single round is 4! = 24. This is because each Westside player can be matched with any Eastside player, and there are 4! ways to do this.However, since each pair needs to play two games, we need to schedule these pairings twice across the 8 rounds. This complicates things because we have to ensure that each pairing occurs exactly twice without overlap in the same round.One approach is to consider that for each round, we have a permutation of the Eastside players. Since we need each pairing to occur twice, we can think of the entire schedule as two copies of a round-robin tournament.In a single round-robin tournament, the number of ways to schedule the matches is (4-1)! = 6 for each Westside player, but since we have four players, it's more complex. Actually, the number of ways to arrange the pairings over multiple rounds is given by the concept of Latin squares or using derangements.But perhaps a simpler way is to think about the total number of possible schedules. Since each round is a permutation of the Eastside players relative to the Westside players, and we have 8 rounds, the total number of possible schedules would be (4!)^8. However, this counts all possible sequences of permutations, including those that don't satisfy the condition of each pair playing exactly two games.To account for the condition that each pair plays exactly two games, we need to divide by the number of ways to arrange the two games for each pair. Since there are 16 pairs (4 Westside × 4 Eastside), and each pair plays two games, the total number of ways to arrange the games is 2^16. However, this might not directly apply because the games are scheduled in rounds.Wait, maybe I'm overcomplicating it. Let's think differently. Each round is a permutation, and we need to choose 8 permutations such that each pair appears exactly twice. This is similar to arranging a timetable where each subject is taught twice a week without overlap.The number of ways to schedule this is given by the number of ways to choose 8 permutations where each pair appears exactly twice. This is a known problem in combinatorics, often related to block designs or scheduling algorithms.Alternatively, we can think of it as arranging the 32 games into 8 rounds of 4 games each, with the constraint that each pair plays exactly two games. The total number of ways to arrange 32 games into 8 rounds is 32! / (4!^8). However, this doesn't account for the pairing constraints.To incorporate the constraints, we need to ensure that in each round, the four games are valid pairings (no two games involve the same player from either school). This is equivalent to ensuring that each round is a derangement or a permutation without fixed points, but actually, it's just a permutation where each player is paired with exactly one opponent.Given that, the number of valid schedules is the number of ways to arrange the 32 games into 8 rounds, with each round being a permutation of the Eastside players relative to the Westside players, and each pair occurring exactly twice.This is a complex combinatorial problem, but I recall that the number of such schedules is given by the formula:Number of schedules = (4!)^8 / (2!^16)This is because for each of the 8 rounds, there are 4! ways to arrange the pairings, but we have to divide by 2! for each pair since each pair plays two games, and the order of the two games doesn't matter.However, this might not be entirely accurate because the two games between each pair are indistinct in terms of scheduling, but the rounds themselves are distinct.Alternatively, another approach is to consider that for each pair, we need to choose 2 rounds out of 8 to play their games. The number of ways to do this for all 16 pairs is:Product over all pairs of C(8,2) = (C(8,2))^16But this counts the number of ways to assign two rounds to each pair, but we also need to ensure that in each round, the four games are valid (no overlapping players).This is similar to a constraint satisfaction problem, and the exact count is non-trivial. However, I recall that the number of such schedules is known and is equal to 8! × (4!)^8 / (2!^16).But I'm not entirely sure about this formula. Let me think again.Each round is a permutation of the Eastside players relative to the Westside players. There are 4! permutations per round, and we have 8 rounds. However, we need to ensure that each pair appears exactly twice. So, it's like arranging 8 permutations such that each pair appears exactly twice.This is equivalent to finding a set of 8 permutations where each pair (i,j) appears exactly twice across the permutations.The number of such sets is given by the number of 8 × 4 matrices where each row is a permutation of 1 to 4, and each pair (i,j) appears exactly twice in the matrix.This is a complex combinatorial object, and I'm not sure of the exact count. However, I recall that the number of such schedules is 8! × 7!! × 6!! × ... × 1!!, but I'm not sure.Alternatively, perhaps it's simpler to think of it as arranging the 32 games into 8 rounds, with each round being a permutation. The total number of ways is 32! / (4!^8), but we need to divide by the number of ways to arrange the games within each round, which is 4! per round, and also account for the fact that the order of the rounds doesn't matter, but in this case, the rounds are distinct because they happen at different times.Wait, actually, the rounds are ordered, so we don't need to divide by the number of rounds. So, the total number of ways is 32! / (4!^8). However, this doesn't account for the pairing constraints.To incorporate the pairing constraints, we need to ensure that each pair plays exactly two games. This is similar to arranging a 4 × 4 biadjacency matrix with exactly two 1s in each row and column, and then counting the number of ways to decompose this matrix into permutation matrices.The number of ways to decompose a 4 × 4 biadjacency matrix with two 1s in each row and column into permutation matrices is given by the number of 2-factorizations of the complete bipartite graph K_{4,4}.The number of 2-factorizations of K_{4,4} is known to be 90. However, since we have two 2-factors (because each pair plays two games), the total number of schedules is 90 × 90 = 8100. But this seems too low.Wait, no, actually, each 2-factorization corresponds to a set of two rounds where each pair plays once. Since we need two games between each pair, we need two 2-factorizations. However, the exact number is more complex.I think I'm getting stuck here. Maybe I should look for a different approach. Let's consider that for each round, we have a permutation of the Eastside players. Over 8 rounds, each pair must appear exactly twice. So, we need to choose 8 permutations such that each pair appears exactly twice.The number of such sets of permutations is equal to the number of 8 × 4 matrices where each row is a permutation of 1 to 4, and each pair (i,j) appears exactly twice in the matrix.This is a type of Latin square, but more complex. I think the number of such matrices is given by the number of ways to arrange the permutations with the given constraints.Alternatively, perhaps it's easier to think about the problem as arranging the games in rounds, ensuring that each pair plays exactly two games. The first round can be any permutation, say, A-W, B-X, C-Y, D-Z. The second round could be a derangement, like A-X, B-Y, C-Z, D-W, and so on.However, ensuring that each pair plays exactly two games across the 8 rounds is non-trivial. I think the number of such schedules is known and is equal to 8! × (4!)^8 / (2!^16). But I'm not sure.Wait, let's think about it differently. Each pair needs to play two games. There are 16 pairs. For each pair, we need to choose 2 rounds out of 8 to play their games. The number of ways to do this for all pairs is (C(8,2))^16. However, this counts the number of ways to assign two rounds to each pair, but we also need to ensure that in each round, the four games are valid (no overlapping players).This is similar to a constraint satisfaction problem, and the exact count is non-trivial. However, I recall that the number of such schedules is given by the formula:Number of schedules = (8! × (4!)^8) / (2!^16)This is because for each of the 8 rounds, there are 4! ways to arrange the pairings, and we have 8! ways to order the rounds. However, since each pair plays two games, we divide by 2! for each pair to account for the indistinguishability of the two games.So, plugging in the numbers:8! = 403204! = 24(4!)^8 = 24^82!^16 = 2^16 = 65536So, the number of schedules is:(40320 × 24^8) / 65536But this seems like a huge number, and I'm not sure if it's correct. Let me check the options given: 30240, 40320, 50400, 60480, 80640.Wait, 40320 is one of the options, which is 8!. Maybe the correct answer is 40320, which is option B.But I'm not entirely confident. Let me think again. If we consider that each round is a permutation, and we need to arrange 8 permutations where each pair appears exactly twice, the number of such schedules is 8! × (4!)^8 / (2!^16). However, this might be overcounting because the two games between each pair are indistinct.Alternatively, perhaps the correct approach is to consider that the number of ways to schedule the games is the number of ways to arrange the 32 games into 8 rounds, with each round being a permutation, and each pair appearing exactly twice.This is equivalent to counting the number of 8 × 4 matrices where each row is a permutation of 1 to 4, and each pair (i,j) appears exactly twice in the matrix.I think the number of such matrices is given by the number of 2-factorizations of K_{4,4}, which is 90, and since we need two 2-factors, the total number is 90 × 90 = 8100. But this doesn't match any of the options.Wait, maybe I'm missing something. The number of 2-factorizations of K_{4,4} is indeed 90, but each 2-factorization corresponds to a set of two rounds where each pair plays once. Since we need two games between each pair, we need two such 2-factorizations. However, the total number of schedules would be the number of ways to arrange these two 2-factorizations over 8 rounds.But this is getting too complicated, and I'm not sure if I'm on the right track. Given the options, and considering that 8! = 40320 is one of them, and it's a common combinatorial result, I think the answer is 40320, which is option B.

💡First, I need to determine the cost of the unknown souvenir type. Let's denote this cost as ( x ) dollars per souvenir.Given that 400 souvenirs of the unknown type were distributed, the total cost for these souvenirs is ( 400x ).The other type of souvenir costs 20 cents each, which is ( 0.20 ) dollars. Since a total of 1000 souvenirs were distributed and 400 were of the unknown type, there must have been ( 1000 - 400 = 600 ) souvenirs of the 20-cent type.The total cost for the 20-cent souvenirs is ( 600 times 0.20 = 120 ) dollars.The total cost of all the souvenirs is given as 220 dollars. Therefore, the cost of the unknown souvenirs plus the cost of the 20-cent souvenirs should equal 220 dollars.This gives us the equation:[400x + 120 = 220]Subtracting 120 from both sides to isolate the term with ( x ):[400x = 100]Dividing both sides by 400 to solve for ( x ):[x = frac{100}{400} = frac{1}{4} = 0.25]So, the unknown type of souvenir costs 25 cents each.

💡First, identify the known values: the diagonal of the closet is 7 feet, and the shorter side is 4 feet.Next, use the Pythagorean theorem to find the length of the longer side. The theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.Set up the equation with the known values:7² = 4² + b²Solve for ( b ):49 = 16 + b²49 - 16 = b²33 = b²b = √33Now that both sides of the closet are known, calculate the area by multiplying the length and width:Area = 4 feet × √33 feetThe area of the closet is ( 4sqrt{33} ) square feet.

💡First, I observe the given pattern:- ( (3 times 2) + 6 = 12 )- ( (4 times 2) + 8 = 16 )- ( (2 times 2) + 10 = 14 )- ( (3 times 2) + 12 = 18 )I notice that each equation involves two steps:1. Multiplying the first number by 2.2. Adding a number that increases by 2 each time to the result of the multiplication.Breaking down the examples:1. ( (3 times 2) + 6 = 6 + 6 = 12 )2. ( (4 times 2) + 8 = 8 + 8 = 16 )3. ( (2 times 2) + 10 = 4 + 10 = 14 )4. ( (3 times 2) + 12 = 6 + 12 = 18 )The number added after the multiplication increases by 2 each time: 6, 8, 10, 12.To find the next element in the sequence starting with ( (1 times 2) ):1. Multiply 1 by 2: ( (1 times 2) = 2 )2. The last number added was 12, so the next number to add would be ( 12 + 2 = 14 ).Finally, add the two results together: ( 2 + 14 = 16 ).Therefore, the next element in the sequence starting with ( (1 times 2) ) is (boxed{16}).

💡First, I will list the factors of each number to identify all possible divisors.For 105, the factors are: 1, 3, 5, 7, 15, 21, 35, 105.For 45, the factors are: 1, 3, 5, 9, 15, 45.Next, I will identify the common factors between the two lists, which are: 1, 3, 5, 15.Among these common factors, the greatest one is 15.Therefore, the greatest common divisor of 105 and 45 is 15.

💡First, I recognize that both 8 and 4 can be expressed as powers of 2. Specifically, 8 is (2^3) and 4 is (2^2).Next, I rewrite the original equation (8^4 = 4^m) using these expressions:[(2^3)^4 = (2^2)^m]Then, I apply the power of a power rule, which states that ((a^b)^c = a^{b times c}):[2^{12} = 2^{2m}]Since the bases are the same, I can set the exponents equal to each other:[12 = 2m]Finally, I solve for (m) by dividing both sides by 2:[m = frac{12}{2} = 6]