T.R | Title | User | Personal Name | Date | Lines |
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1432.1 | marginal problem | HERON::BUCHANAN | Holdfast is the only dog, my duck. | Mon Apr 29 1991 11:40 | 6 |
| Ha! A quaint setting.
Now, what about the same problem, *without* replacement?
Cheers,
Andrew.
|
1432.2 | | CADSYS::COOPER | Topher Cooper | Mon Apr 29 1991 15:53 | 4 |
| I have an interesting proof for my answer, but I'm afraid there isn't
room in this note for it.
Topher
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1432.3 | is a countable infinity of answers enough? | CSSE::NEILSEN | Wally Neilsen-Steinhardt | Mon Apr 29 1991 16:10 | 3 |
| > Now, what about the same problem, *without* replacement?
k > 2, #black < k, #white < k, #red < k
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1432.4 | solved for k =< 2 | HERON::BUCHANAN | Holdfast is the only dog, my duck. | Mon Apr 29 1991 16:47 | 34 |
1432.5 | k=3 has many solutions | CLT::TRACE::GILBERT | Ownership Obligates | Tue Apr 30 1991 13:25 | 20 |
1432.6 | k=4 | CLT::TRACE::GILBERT | Ownership Obligates | Tue Apr 30 1991 13:43 | 3 |
| And b!/(b-4)! + w!/(w-4)! = r!/(r-4)!
has solutions:
[7,7,8], and [132,190,200]
|
1432.7 | quick remark | HERON::BUCHANAN | Holdfast is the only dog, my duck. | Wed May 01 1991 11:15 | 6 |
| In fact there's at least 1 solution for each k > 0.
Let w = b = 2k-1, r = 2k.
Cheers,
Andrew.
|