answer:To solve this problem, we can use the concept of permutations. Since there are 8 candidates and we need to choose 1 president, 1 vice president, and 3 members, we can calculate the number of ways to choose the council as follows: 1. Choose the president: There are 8 candidates, so there are 8 ways to choose the president. 2. Choose the vice president: After choosing the president, there are 7 candidates left, so there are 7 ways to choose the vice president. 3. Choose the 3 members: After choosing the president and vice president, there are 6 candidates left. We need to choose 3 members from these 6 candidates, which can be done in 6!/(3!*(6-3)!) ways using the binomial coefficient formula. So, the total number of ways to choose the council is: 8 (ways to choose president) * 7 (ways to choose vice president) * (6!/(3!*3!)) (ways to choose 3 members) 8 * 7 * (6!/(3!*3!)) = 8 * 7 * (720/(6*6)) = 8 * 7 * 20 = 1120 There are 1120 ways to choose the student council using Multinomial coefficients.

answer:To divide the 6 different objects into 3 groups of 2 objects each, we can use the concept of combinations. First, we choose 2 objects out of 6 for the first group. The number of ways to do this is C(6,2) = 6! / (2! * (6-2)!) = 15. Next, we choose 2 objects out of the remaining 4 for the second group. The number of ways to do this is C(4,2) = 4! / (2! * (4-2)!) = 6. Finally, the last 2 objects will automatically form the third group. Now, we multiply the number of ways to form each group: 15 * 6 = 90. However, since the groups are indistinguishable, we have overcounted the arrangements. We need to divide by the number of ways to arrange the 3 groups, which is 3! = 6. So, the final answer is 90 / 6 = 15 different ways to divide the 6 objects into 3 groups of 2 objects each.

answer:To ensure that each language group is represented by at least one student, the teacher must select one student from each of the French, Spanish, and German groups, and then choose the fourth student from the remaining students. First, the teacher can choose one French student in 6 ways (since there are 6 French students), one Spanish student in 5 ways (since there are 5 Spanish students), and one German student in 9 ways (since there are 9 German students). So far, there are 6 * 5 * 9 = 270 ways to choose one student from each language group. Now, the teacher needs to choose the fourth student from the remaining students. Since the teacher has already chosen one student from each language group, there are now 19 students left to choose from (20 total students - 1 French - 1 Spanish - 1 German = 19). The teacher can choose the fourth student in 19 ways. Therefore, the total number of ways the teacher can select a committee of 4 students with each language group represented by at least one student is 270 * 19 = 5130 ways.

answer:We can solve this problem using the Inclusion-Exclusion Principle. First, let's find the total number of ways to distribute the books without any restrictions. There are 3 students, and each book can be given to any of the 3 students. So, for each book, there are 3 choices. Since there are 5 books, the total number of ways to distribute the books is 3^5 = 243. Now, we need to subtract the cases where at least one student doesn't receive a book. There are 3 ways to choose which student doesn't receive a book, and then there are 2 choices for each of the 5 books (either given to the first student or the second student). So, there are 3 * 2^5 = 96 ways where at least one student doesn't receive a book. However, we have subtracted the cases where two students don't receive a book twice, so we need to add those cases back. There are 3 ways to choose which student receives all the books, and there is only 1 way to distribute the books in this case (all 5 books go to the chosen student). So, there are 3 * 1 = 3 ways where two students don't receive a book. Using the Inclusion-Exclusion Principle, the total number of ways to distribute the books such that each student receives at least one book is: Total ways - Ways with at least one student without a book + Ways with two students without a book = 243 - 96 + 3 = 150. So, there are 150 ways to distribute the five different math textbooks to three students, ensuring that each student receives at least one book.