what is the formula to use here? how many groups can be formed with a, a, a, a, b, b, b, c, c? the groups can comprise any number of members. thanks.

Mozartain wrote:what is the formula to use here? how many groups can be formed with a, a, a, a, b, b, b, c, c? the groups can comprise any number of members. thanks. 4 a 3 b 2 c 1 member a b c 2 members aa bb cc ab ac cb 3 members abb acc aab aac abc aaa bcc bbc bbb 4 members aaaa aaab aaac aabb aacc aabc abbb abbc abcc accc bbbc 5 members BLA BLA BLA 6 members BLA BLA BLA 7 members aaaabbbcc Somebody help me!!

Here is how I went about it and what I got... There are 9 members in the group and the subgroups are unordered. The subgroups can be any number of members. So, for each number of the members of the subgroups, you need to find the number of groups there and then add them all up. In other words - how many group with 1 members, 2 members, 3 members - 9members and the sum of all these possibilities will be the answer to "How many subgroups of any size can be formed from a 9 member group?" For Members=1: 9! / 1!!*9! = 9 M=2: 9! / 2!*8! = 36 M=3: 9!/3!*6! = 84 M=4: 9!/4!*5! = 126 Now note that the formulas for M=5-8 are the same as M=4-1, so you have 2 (9+36+84+126) = 355 gets you all the possible combinations for subgroups size 1-8. By the way the original question is worded, it is hard to know if you need this extra step or not (although from what I have seen of GMAT questions, I would expect so)...remember that there is one more group, that is the one composed of all 9 members (just 1 combination), so the total groups of any numbers of members (1-9 inclusive) is 356. -Carrie

Logitech, I didn't see anything in the question that said the subgroups couldn't be comprised of members of the same type (a b or c) or that the types had any effect on the subgroups at all. Hence, I just counted the 9 members and went from there. Perhaps the question wording could be more clear? -Carrie