SOLUTION: In how many ways can a group of 6 people be divided into two unequal groups if there must be at least one person in each group? I've tried combinations formula but it didnt work pl

You may have these two basic configurations:


          Table

     1-st group    2-nd group
---------------------------------

1    1 persons     5 persons

2    2 persons     4 persons


And, basically, that's all.


Now we need to calculate in how many ways you can collect each configuration.


a)  Configuration #1:  1 person in one group and 5 persons in other group.


    You cam choose one person of 6 by 6 ways.

    In this way, as soon as you fill the first group, the second group is formed AUTOMATICALLY of 5 remained persons.

    So, you have 6 options/opportunities to make the configuration #1.



b)  Configuration #2:  2 persons in one group and 4 persons in other group.


    You can choose two persons of 6 by   =  = 15 ways.

    Working on this configuration, as soon as you fill the first group of 2, the second group of 4 is formed AUTOMATICALLY of 4 remained persons.

    So, you have 15 ways/options/opportunities to make the configuration #2.


c)  Since your options in Case #1  and in Case #2 are independent, you have 6*15 = 90 ways, in all, to make your selections.