Seating Arrangement: Boys And Girls On A Bench

Hey guys! Ever wondered how many different ways a group of people can sit together, especially when there are some rules? Let's dive into a fun math problem about seating arrangements. This problem involves calculating the number of possible seating arrangements for four boys and four girls on a bench, with the condition that boys must occupy the ends of the bench. Sounds interesting, right? Let's break it down step by step!

Understanding the Problem

So, we have four boys and four girls who need to be seated on a bench. The catch here is that the two ends of the bench must be occupied by boys. This constraint makes the problem a bit more interesting. To solve this, we'll use the principles of permutations, which is a way of arranging things in a specific order. Permutations are crucial in understanding how many different arrangements are possible given certain conditions. Remember, the order matters in permutations, so ABC is different from BCA.

Breaking Down the Constraints

First, let’s focus on the boys. We have four boys, and we need to choose two of them to sit at the ends of the bench. The order in which they sit matters because if Boy A sits on the left and Boy B on the right, it's a different arrangement than Boy B on the left and Boy A on the right. This is a classic permutation scenario. We need to figure out how many ways we can arrange two boys out of four for these end positions. This initial step is critical because it sets the stage for the rest of the seating arrangement. Getting this part right ensures we adhere to the main condition of the problem.

Calculating End Positions

To calculate the number of ways to arrange the boys at the ends, we use the permutation formula. The formula for permutations is P(n, r) = n! / (n - r)!, where n is the total number of items, and r is the number of items we are choosing and arranging. In our case, n = 4 (the number of boys) and r = 2 (the number of end positions). So, we have P(4, 2) = 4! / (4 - 2)! = 4! / 2! = (4 × 3 × 2 × 1) / (2 × 1) = 12. This means there are 12 different ways to seat two boys at the ends of the bench. This calculation is a cornerstone of solving the problem, providing a solid foundation for the next steps. Think of it as setting the stage for a play; once the main actors are in place, the rest can follow.

Arranging the Remaining Students

Now that we've seated the boys at the ends, let’s tackle the rest of the group. After placing two boys at the ends, we have two boys and four girls remaining, making a total of six students to be seated. These six students can sit in any order in the remaining six seats. The number of ways to arrange these six students is a permutation of six items taken six at a time, which is simply 6! (6 factorial). This part of the problem focuses on the flexibility within the remaining seats, allowing for different combinations of boys and girls. It’s like filling in the middle of a sandwich after the bread (the boys at the ends) is set.

Calculating the Arrangements

To find 6!, we multiply all positive integers up to 6: 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720. This means there are 720 different ways to arrange the remaining six students in the middle seats. This calculation is straightforward but essential, as it quantifies the variety of arrangements possible in the core section of the seating. Each of these 720 arrangements can be combined with the end arrangements, significantly increasing the total number of seating possibilities.

Combining the Possibilities

To get the total number of seating arrangements, we need to combine the number of ways to arrange the boys at the ends with the number of ways to arrange the remaining students. We do this by multiplying the two results together. Remember, for each arrangement of the boys at the ends, there are 720 different arrangements for the middle six students. So, we multiply 12 (ways to arrange boys at the ends) by 720 (ways to arrange the remaining students). This step is crucial because it synthesizes the individual calculations into the final answer. It’s like putting the ingredients together to bake a cake – each part is important, but the final product is what we’re after.

Final Calculation

Let's do the final calculation: 12 × 720 = 8640. Therefore, there are 8,640 possible seating arrangements where boys occupy the ends of the bench. This is our final answer, representing the total number of different ways the students can be seated under the given conditions. This number highlights the power of permutations in solving combinatorial problems. It’s not just about math; it’s about understanding how many options we have in arranging things, which has applications in various fields, from scheduling to cryptography.

Final Answer

So, the final answer is 8,640 possible arrangements. Wasn't that a fun problem to solve? By breaking it down into smaller parts, we could easily tackle the constraints and calculate the total number of possibilities. Remember, when you encounter similar problems, always start by identifying the constraints and then use permutations or combinations to find the answer. Keep practicing, and you'll become a pro at solving these types of questions! And that's it for this problem, guys! Hope you enjoyed the breakdown. Keep your thinking caps on, and see you in the next math adventure!

Answer

B. 8.640 susunan