Re: Puzzle

*To*: mathgroup at smc.vnet.net*Subject*: [mg2294] Re: Puzzle*From*: bshoels at rs1.tcs.tulane.edu (Brett Shoelson)*Date*: Mon, 23 Oct 1995 12:41:13 -0400*Organization*: Tulane University

Will Self (wself at viking.emcmt.edu) wrote: : Here's a nice puzzle. Find the four numbers in the following sequence : which total 100: {6, 44, 30, 15, 24, 12, 33, 23, 18}. It would seem that : there might be a nice way to do this with Mathematica. Any takers? : Will Self : Billings, Montana There are 126 possible 4-number combinations from the set of 9 numbers... (9!/(4! * (9-4)!). How to generate these sets is another matter entirely. Unless I am missing something, there seems to be no Combinations[list,n] function which gives all combinations of length n from list. I could write a recursive Table command to generate them, but it would not be a very elegant approach. Once the set of all 4-number sets is generated, the problem becomes easy in Mma. Here is one approach, which, in my example, selects the subsets of Example which sum to 17: Example = {{6,7,4},{3,5,6},{1,2,7},{9,3,5}}; sums = Table[Apply[Plus,Example[[i]]],{i,Length[Example]}]; positions = Flatten[Position[MySums,17]]; Table[Example[[positions[[i]]]],{i,1,Length[positions]}] {{6, 7, 4}, {9, 3, 5}} So, the problem is ESSENTIALLY solved. Now, if anyone can tell me how to (easily) generate the set of 4-number combinations (subsets) of the original 9-member set, I would be very grateful. Brett Shoelson bshoels at rs6000.tcs.tulane.edu