Question

I have four identical oranges. How many ways are there for me to divide these oranges into at most three groups? (By definition, a group must have at least one orange.)

Answer 1

Group 1 - all groups of 1 orange. There are only 4 oranges, so 4 groups
Group 2 - all groups of 2 oranges. 4 oranges arranged in groups of 2 is 4 choose 2
Group 3 - all groups of 3 oranges. 4 oranges arranged in groups of 3 is the same as 4 oranges arranged in groups of 1
Add all these 3 possibilities.

Answer 2

Well, wouldn't group 1 not count? Because it says at most 3 groups?

Answer 3

Also, I think it means to put them into groups simultaneously, and with no more than 3 groups

Answer 4

There are not many ways to make groups. Only group 4 is left.

Answer 5

@satellite73 If you have time, can you help me on this one?

Answer 6

dunno, too many glasses of wine
but we can try it
1 group, 1 way
two groups i would say 4 ways, but that might be wrong because it says you cannot tell them apart

Answer 7

lol

Answer 8

three groups i would say \(4\times 3\times 2\times 3\) but i am confused about telling them apart

Answer 9

Okay, I've confirmed that it rearrangements of the same way are still considered the same.
i.e. 1,3 3,1 would count as 1 way, not two.

Answer 10

http://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

Answer 11

theorem two tells us \(\dbinom{3+4-1}4=\dbinom64=\dfrac{6!}{2!4!}=15\)

Answer 12

How did you get 3 and 4? I don't quite understand the second theorem.

Answer 13

oh permutations are identical?

Answer 14

Yes

Answer 15

Which is why I'm thinking it is just-
1 Group- 4 balls all together, 1 way
2 Groups- 1,3 (3,1) 2,2 (2,2) 2 ways
3 Groups - 1,1,2 (2,1,2;2,1,1) 1 way....

Am I missing something?

Answer 16

@wgary essentially we're counting the number of ways we can place 'bars' to separate the oranges: