Combining Objects in Buckets

Apr 2012
5
0
Wisconsin
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?
 
Last edited:
Apr 2012
5
0
Wisconsin
Is this one without an easy, mathematical solution?
 
Dec 2007
3,184
558
Ottawa, Canada
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 ?
 
Last edited:
Apr 2012
5
0
Wisconsin
Inclusively. {0, 1, 2, 3, and 4}

And, whups! I meant G and H cannot be taken together.
 
Dec 2007
3,184
558
Ottawa, Canada
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?
 
Apr 2012
5
0
Wisconsin
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.
 
Dec 2007
3,184
558
Ottawa, Canada
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:D
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?
 
Apr 2012
5
0
Wisconsin
There are ONLY 4 of these; I show:
DFH:B ; also possible are:
BFH:D
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.
 
Dec 2007
3,184
558
Ottawa, Canada
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)