Is this one without an easy, mathematical solution?
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?
HOKAY! Une autre question:
on this 4:3:1 arrangement :
There are ONLY 4 of these; I show:Code:1 2 3 4 5 6 A x B x C x D x E x F x G x H x
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?
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"