Một nhóm gồm 10 học sinh, trong đó có 7 nam và 3 nữ. Hỏi có bao nhiêu cách sắp xếp 10 học sinh trên thành một hàng ngang sao cho 2 vị trí đầu cuối là 2 em nam và không có 2 em nữ nào ngồi cạnh nhau

To solve this problem, we can use the principle of counting and permutation. First, let's consider the number of ways to arrange the 7 boys and 3 girls without any restrictions. This can be calculated using the permutation formula: $$ Where n is the total number of students and r is the number of students to be arranged. So, the number of ways to arrange 10 students without any restrictions is: $$ Next, we need to consider the restriction that two boys must occupy the first and last positions. After fixing these two positions for boys, we have 8 remaining positions for arranging the other students. Now, let's consider the number of ways to arrange the 3 girls among these 8 positions. We can calculate this using permutation as well: $$ Finally, we need to account for the restriction that no two girls can sit next to each other. To do this, we'll use the principle of inclusion-exclusion. Let A be the event that two or more girls sit next to each other. Then we want to find $$, which represents all arrangements where no two girls sit next to each other. We can calculate $$ by subtracting from $$ all arrangements where at least two girls sit together. This can be expressed as: $$ The arrangements with exactly 2 adjacent girls can be calculated by treating them as a single entity (GG), so there are actually only 7 entities (6 positions between boys and one at each end). This gives us: $$ Similarly, for arrangements with exactly 3 adjacent girls: $$ Substitute these values into our equation for $$: $$ After calculating $$, we get a large negative value which represents the number of valid arrangements satisfying all given conditions. Therefore, Number of valid arrangements: -32510419200