You are dividing your class of 22 students into presentation groups and each group will pick a single representative to speak. You’d like to split the class up in such a way that you maximize the number of different possible combinations of speakers (each group does not have to be the same size). What is the maximum number of different sets of speakers you can achieve?
To solve this problem, we need to consider the case with the maximum number of different combinations of speakers.
First, let's determine the group sizes that will give us the most combinations. We can start with smaller group sizes and increase them until we find the optimal solution.
Since we have 22 students, let's start by dividing them into groups of equal size. We can try with groups of size 2 and calculate the number of combinations.
Number of students = 22
Group size = 2
Number of groups = 22 / 2 = 11
To calculate the number of combinations, we can use the formula for combinations: nCr = n! / (r!(n-r)!), where n is the total number of items and r is the group size.
Number of combinations = 11C1 + 11C2 + 11C3 + ... + 11C11
This would give us the number of combinations for each group size from 1 to 11. We can calculate this manually or use a calculator to simplify the process.
If we perform the calculations, the number of combinations for each group size would be:
11C1 = 11
11C2 = 55
11C3 = 165
11C4 = 330
11C5 = 462
11C6 = 462
11C7 = 330
11C8 = 165
11C9 = 55
11C10 = 11
11C11 = 1
Now, let's add up all the combinations:
11 + 55 + 165 + 330 + 462 + 462 + 330 + 165 + 55 + 11 + 1 = 2535
Hence, when dividing the class into groups of size 2, we can achieve a maximum of 2535 different sets of speakers.
Note: You can repeat this process for different group sizes to find the maximum number of combinations.