20060421, 07:58  #1 
Dec 2005
2^{2}×7^{2} Posts 
Combinatorics
You have a set of k!+k1 elements (k>1), numbered 1, 2,..., k!+k1 .
Let A be the set of subsets with exactly k elements. Let B be the set of subsets with exactly k1 elements, which includes also all the permutations of such a subset (so if for instance k=4 then the subsets {1,2,3},{1,3,2},{2,3,1},{2,1,3},{3,1,2}{3,2,1} are all in B. Evidently #A=#B Can you construct a bijection f such that f(A_i)=B_i and such that for all i A_i contains B_i ? 
20060503, 11:14  #2 
Dec 2005
2^{2}·7^{2} Posts 
In all honesty, for me it is an open question

20060504, 01:02  #3 
Jun 2004
Chicago
2^{2}×7 Posts 
Given a kelement subset, remove any of the k elements, this clearly is a permutation of k1 elements.
Take a set (k1)element subsets of the same k1 elements. There are k! remaining elements that can be placed back into the (k1)element subsets. Now, there are exactly k repeated kelement subsets doing this process, so there are only (k1)! that can be added to each set of permutations. Now there are (k1)! permutations, and (k1)! elements that can be added to those permutations, adding an element to each permutation out of the possible (k1)! elements gives a distinct kelement subset. Therefore, bijection.  is that what you were looking for, or am I off? Last fiddled with by jebeagles on 20060504 at 01:04 
20060509, 08:38  #4 
Dec 2005
2^{2}·7^{2} Posts 
At least you are trying for a solution. I wonder however whether I should accept this solution. It seems all right, but it silently assumes the following:
if during the process of 'adding' an element out of the (k1)! available to one of the (k1)! permutations we get a ksubset twice, this would mean that another ksubset does not get attributed and we should be able to settle that. There is yet another snag: for the moment we have only looked at one subset of k1 elements.There are however quite some subsets of k1 elements. Are we still getting distinct ksubsets ? 
20060509, 11:53  #5 
Dec 2005
2^{2}·7^{2} Posts 
you are right, the second part of my answer is gibberish at best...
what jebeagles is showing is the following: we have the set {1,...,k!+k1} and we take some subset of k1 elements. say {1,...,k1}. To this subset we can 'add' k! different elements. Here we have our first problem. The number of sets {1,...,k1} including permutations is (k1)! How to add the remaining elements? (we had k!=(k1)! * k elements remaining after selecting k1 elements). So which do we select ? And what happens to the selectionprocedures for other subsets (for instance {1,...,k2,k}) after this selection. Can we still fill. I am not really satisfied with the answer, actually. All that seems to be shown is that the two sets have the same size (which I offered at the beginning of the problem). If that was the end to it, we would obviously have a bijection. But the extra condition makes this unclear for me. 
20060512, 19:25  #6 
Aug 2003
Upstate NY, USA
2×163 Posts 
If you number the subsets by ranking from least to greatest for first term, then second term, then third term, then fourth
i.e. A1={1,2,3,4} A2={1,2,4,3} A3={1,3,2,4} A4={1,3,4,2} ... and f({a,b,c,d})={a,b,c} then B1={1,2,3} B2={1,2,4} B3={1,3,2} B4={1,3,4} ... which I believe meets your requirements, but I may be reading something incorrectly 
20060513, 06:58  #7 
Dec 2005
11000100_{2} Posts 
The problem is that the subsets in A do not have permutations of themselves
A1=(1,2,3,4), A2=(1,2,3,5) B1=(1,2,3), B2=(1,3,2) 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Combinatorics news  Nick  Combinatorics & Combinatorial Number Theory  7  20191214 15:25 
Stacking boxes  combinatorics?  Ken_g6  Puzzles  15  20100804 23:30 