Final answer:
There are 15 different ways to divide 6 people into 3 groups of 2 when considering the groups distinct from each other. This can be calculated using combinations and accounting for the number of ways to arrange the groups.
Step-by-step explanation:
To divide 6 people into 3 groups, we need to consider the different ways we can partition the group. These groups can be of different sizes, and each specific arrangement where the order of groups matters is a different outcome.
However, this problem doesn't specify sizes for the groups, but implies equal groups. Therefore each group would have 2 people, and we are trying to find the number of unique sets of pairs. We can think of this as choosing 2 people for the first group, then 2 for the second group out of the remaining, and the last 2 people automatically forming the third group.
To calculate this, we could use the combination formula which is nCr = n! / [r! * (n-r)!]. Here's how we use the formula:
- Choose 2 people out of 6 for the first group: 6C2 = 15.
- Choose 2 people out of the remaining 4 for the second group: 4C2 = 6.
- The last 2 people form the third group.
To find the total number of ways to create the groups, we multiply these combinations, but since the order of groups doesn't matter, we need to divide by the number of ways to order 3 groups, which is 3! (3 factorial).
Total different ways = (6C2 * 4C2) / 3! = (15 * 6) / 6 = 15 ways
However, for 6 people divided into 3 groups of 2, there are 15 different ways to do this if we consider the groups distinct from each other.