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Re: Combination Algorithm without brut force - Combine 4's into least 6's

• To: mathgroup at smc.vnet.net
• Subject: [mg27287] Re: [mg27284] Combination Algorithm without brut force - Combine 4's into least 6's
• From: BobHanlon at aol.com
• Date: Sat, 17 Feb 2001 03:31:06 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

Needs["DiscreteMath`Combinatorica`"];

Use KSubsets

Length[KSubsets[Range[10], 4]] == Binomial[10, 4] == 210

True

I have no idea of what you mean by "combining all 4s into the least number
of 6s"


Bob Hanlon

In a message dated 2001/2/16 4:43:29 AM, aufempen at modemss.brisnet.org.au 
writes:

>I have Mathematica 2.2 for WIN3.1 and I have not got much combinatorial
>functions on it.
>Could you please help?
>May be I should post this on the Wolfram forum? which one?
>
>What is the algorithm  or principle to combine all the combinations of
>4s
>into
>the smallest combination of 6s from a range of 10 consecutive numbers?
>What is the combinatorix formula for the least combination of 6s?
>Here is the data
>1) All the combinations of 4s in a range of 10 numbers
>  1           1  2  3  4
> 2             1  2  3  5
> 3             1  2  3  6
>....
>The full result of all combinations of 4s has
> been truncated to save space on this postingl
>......
> 208           6  7  9  10
> 209           6  8  9  10
> 210           7  8  9  10
>
>2) The result for combining all 4s into the least number
>of 6s is 21 combinations of 6s:
>But how do you set the algorithm or explain how it works
>without using brut force?
>1          1  2  3  4  5  6
>2          1  2  3  4  7  8
>3          1  2  3  4  9  10
>4          1  2  4  5  7  10
>5          1  2  4  6  8  9
>6          1  2  5  6  7  9
>7          1  2  5  6  8  10
>8          1  3  5  6  7  8
>9          1  3  5  6  9  10
>10       1  3  7  8  9  10
>11       1  4  5  8  9  10
>12       1  4  6  7  9  10
>13       2  3  4  5  8  10
>14       2  3  4  6  7  9
>15       2  3  5  7  9  10
>16       2  3  6  8  9  10
>17       2  4  5  7  8  9
>18       2  4  6  7  8  10
>19       3  4  5  6  7  10
>20       3  4  5  6  8  9
>21       5  6  7  8  9  10
>
>Brut force = Obtaining these numbers by designing my computer program
>to number crunched all possibilities.
>
>3)Any explanation, containing combination, permutation, group mathematics
>will
>be appreciated. Please post to aufempen at modemmss.brisnet.org.au
>