At a boarding school, in the evening, 9 students always take walks in groups of 3. How can the groups be arranged so that each person walks in a group with every other person exactly once in 4 days?
I would like to know whether there is a trick to these type of questions. My workings are students numbered 1-9 and I did 123 124 125 126 127 128 129. And I would also like to know whether my answer could be correct, 9 students, 3 spots is 27 different combinations?
To arrange the groups so that each person walks with every other person exactly once in 4 days, you can use a rotating schedule. Here is one possible way to arrange the groups:
Day 1:
Group 1: 1, 2, 3
Group 2: 4, 5, 6
Group 3: 7, 8, 9
Day 2:
Group 1: 1, 4, 7
Group 2: 2, 5, 8
Group 3: 3, 6, 9
Day 3:
Group 1: 1, 5, 9
Group 2: 2, 6, 7
Group 3: 3, 4, 8
Day 4:
Group 1: 1, 6, 8
Group 2: 2, 4, 9
Group 3: 3, 5, 7
Each person will walk with every other person exactly once in these four days.
Regarding your question about the number of combinations, you are correct that there are 27 different combinations if we consider 9 students and groups of 3. Each group is considered unique, so there would be 27 different possible combinations of groups.
To solve this question, we need to arrange the students into groups of three so that each student walks with every other student exactly once in a span of four days.
First, let's break down the problem into smaller steps:
1. Calculate the total number of possible groups:
Since there are 9 students and each group has 3 students, we can use the combination formula to find the total number of possible groups. The formula is nCr = n! / (r!(n-r)!), where n is the total number of students and r is the number of students in each group.
In this case, n = 9 and r = 3, so the total number of possible groups is 9C3 = (9! / (3! * (9-3)!)) = 84.
2. Find the number of unique combinations possible:
Since each student needs to walk with every other student exactly once in four days, we need to determine how many unique combinations we can create given the total number of possible groups.
To calculate this, we can use the formula nP2 = n! / (n-2)!, where n is the total number of students.
In this case, n = 9, so the number of unique combinations is 9P2 = (9! / (9-2)!) = 9! / 7! = 72.
Now, let's check if your answer is correct:
You said that you arranged the groups as follows: 123 124 125 126 127 128 129. This gives us a total of 7 groups.
To determine if this arrangement is correct, let's count the number of unique combinations:
Within each group, there are 3C2 combinations that can be formed (since each student walks with every other student once). Thus, from each group, we get 3 combinations.
Therefore, the total number of unique combinations from your arrangement would be 7 groups * 3 combinations = 21 combinations, which is different from the expected 72 combinations.
Hence, it seems that your arrangement does not fulfill the requirement of each student walking with every other student exactly once in four days.
There might be additional valid arrangements that satisfy the given conditions. One way to solve this problem is to use graph theory, specifically the concept of "complete graphs" or "cliques." By constructing a complete graph with 9 vertices (representing the students) and labeling the edges with the days, you can find the required arrangements.
In complex problems like this, it can be helpful to break down the problem into smaller steps and use mathematical concepts to find the solution.