Hi - I know that groups of 2 is not an answer choice for this question, but I wanted to understand why that would not give the greatest amount of options? Following the combinatorics we would do 8!/(2!*4) to represent each of sets of 2 students, which equals 5040 (much greater than the 70 we found for the answer). Could someone tell me where I’m going wrong with my logic? Thanks!
If you think about 8 people and they are put in to groups of 2, 1 student can only work in 7 distinct groups (Student 1 with Student 2, 3, 4, 5, 6, 7 or 8 or 7!/((1!)(6!)). There are 5040 combinations to pair and order these groups but from the perspective of each student, they will only ever be in 7 different pairs.
If Student 1 has a possible 3 other students to be in a group with they then have 7!/((3!)(4!) or 35 distinct groups they can work in. If the groups can be ordered (i.e. Group 1 or 2) then this makes the 70 permutations found in the question.
That makes sense! Thank you!