# Combining Objects in Buckets

#### banantalis

Howdy!

I have a set of 8 objects: {A, B, C, D, E, F, G, H}
I have 6 buckets.
I must place all 8 objects in any combination of the 6 buckets.

I can place between 0 and 4 objects in any bucket.
I must place all objects in a bucket.

A and B cannot be placed in the same bucket.
C and D cannot be placed in the same bucket.
E and F cannot be placed in the same bucket.
F and G cannot be placed in the same bucket.

The order of objects in the bucket is irrelevant. (Bucket 1: ACEF is identical to Bucket 1: FECA, i.e.)
The order of buckets is irrelevant. (Bucket 1: ACEF, Bucket 2: BDEG is identical to Bucket 1: BDEG, Bucket 2: ACEF is identical to Bucket 4: BDEG, Bucket 6: ACEF, i.e.)

The question is: how many arrangements exist? How do I figure this one out?

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#### banantalis

Is this one without an easy, mathematical solution?

#### Wilmer

I can place between 0 and 4 objects in any bucket.
0 and 4 inclusive (meaning 5 choices: 0,1,2,3,4) or "between" 0 and 4 (meaning 3 choices: 1,2,3)?

AND:
"F and G cannot be placed in the same bucket."
Is that correct, or do you mean G and H ?

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#### banantalis

Inclusively. {0, 1, 2, 3, and 4}

And, whups! I meant G and H cannot be taken together.

#### Wilmer

Soooo....because of order of buckets and objects being irrelevant,
there is ONLY 1 way to have 2 buckets at maximum (containing 4) :
A,C,E,G in any bucket, B,D,F,H in any of remaining 5 buckets....is that your intent?

#### banantalis

Soooo....because of order of buckets and objects being irrelevant,
there is ONLY 1 way to have 2 buckets at maximum (containing 4) :
A,C,E,G in any bucket, B,D,F,H in any of remaining 5 buckets....is that your intent?
Yes.

#### Wilmer

HOKAY! Une autre question:
on this 4:3:1 arrangement :
Code:
   1   2   3   4   5   6
A  x
B          x
C  x
D      x
E  x
F      x
G  x
H      x
There are ONLY 4 of these; I show:
DFH:B ; also possible are:
BFH
BDH:F
BDH:H
But there is NO MORE using these 4 letters.
Of course, same can be done with A,C,E,G.

So there are eight 4:3:1 arrangements.

Oui?

#### banantalis

There are ONLY 4 of these; I show:
DFH:B ; also possible are:
BFH
BDH:F
BDH:H
But there is NO MORE using these 4 letters.
Of course, same can be done with A,C,E,G.

So there are eight 4:3:1 arrangements.

Oui?
By my understanding of the problem, that looks right. It would be BDF:H, but yes.

#### Wilmer

By my understanding of the problem, that looks right.
Well, pretty "discouraging" if YOU are not completely sure what the problem means.

You answered "yes" to this question earlier:
Soooo....because of order of buckets and objects being irrelevant,
there is ONLY 1 way to have 2 buckets at maximum (containing 4) :
A,C,E,G in any bucket, B,D,F,H in any of remaining 5 buckets....is that your intent?

However, I goofed: like, A,D,E,G with B,C,F,H also possible...along with another 6 similar cases.

Here's the 8 cases (I'll assign values; A=1, B=2, .... , H=8):
1357,2468
1358,2467
1367,2458
1368,2457
1457,2368
1458,2367
1467,2358
1468,2357

So, there are 8 ways if 2 buckets are used, not only 1.
In other words, 8 combinations where 12,34,56,78 are NOT seen in any bucket!

Next step is to do similar work using 3 buckets...then 4 buckets...keep going.

But I'm not doing no more...somebody else's turn to have his/her head "spin" (Wink)