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let-S-a-b-c-d-e-f-if-we-take-any-subset-S-same-subset-is-allowed-it-also-can-be-S-which-will-form-S-if-we-join-them-order-of-operation-does-not-matter-a-b-c-d-d-e-f-is-the-same-as-




Question Number 190602 by uchihayahia last updated on 07/Apr/23
      let S={a,b,c,d,e,f}   if we take any subset S (same subset is allowed),   it also can be S, which will form S if we join them,  order of operation does not matter   ({a,b,c,d},{d,e,f}) is the same as   ({d,e,f},{a,b,c,d})   how many ways can we choose?
letS={a,b,c,d,e,f}ifwetakeanysubsetS(samesubsetisallowed),italsocanbeS,whichwillformSifwejointhem,orderofoperationdoesnotmatter({a,b,c,d},{d,e,f})isthesameas({d,e,f},{a,b,c,d})howmanywayscanwechoose?
Answered by mr W last updated on 07/Apr/23
 { (1),(6) :} {: (),() }+ { (2),(6) :} {: (),() }+ { (3),(6) :} {: (),() }+ { (4),(6) :} {: (),() }+ { (5),(6) :} {: (),() }+ { (6),(6) :} {: (),() }  =1+31+90+65+15+1  =203  (=B_6 =203)  with   { (k),(n) :} {: (),() }=S(n,k)=Stirling numbers of 2. kind  B_n =Bell numbers
{16}+{26}+{36}+{46}+{56}+{66}=1+31+90+65+15+1=203(=B6=203)with{kn}=S(n,k)=Stirlingnumbersof2.kindBn=Bellnumbers
Commented by uchihayahia last updated on 07/Apr/23
for example   subset with 0 element    and subset with 6 elements there is 1 choice   subset with 1 element   and subset with 5 elements there is 6 choices   subset with 1 element   and subset with 6 elements there is 6 choices   subset with 2 element   and subset with 6 elements there is 15 choices   subset with 2 element   and subset with 5 elements there is 30 choices   subset with 2 element   and subset with 4 elements there is 15 choices    so on and so forth
forexamplesubsetwith0elementandsubsetwith6elementsthereis1choicesubsetwith1elementandsubsetwith5elementsthereis6choicessubsetwith1elementandsubsetwith6elementsthereis6choicessubsetwith2elementandsubsetwith6elementsthereis15choicessubsetwith2elementandsubsetwith5elementsthereis30choicessubsetwith2elementandsubsetwith4elementsthereis15choicessoonandsoforth
Commented by mr W last updated on 07/Apr/23
Commented by uchihayahia last updated on 07/Apr/23
 the subset can be empty set, also   any subset taken doesn′t have to disjoint.   i think it′s more than 52. i′m sorry if   my qusetion was not clear
thesubsetcanbeemptyset,alsoanysubsettakendoesnthavetodisjoint.ithinkitsmorethan52.imsorryifmyqusetionwasnotclear
Commented by mr W last updated on 07/Apr/23
i′m sorry that it is still not clear to  me. can you explain with some   examples what is not included in  my solution and what you mean  with empty subsets?  btw: i misread that the set has only  5 elements. but it has 6 elements.  this is fixed.
imsorrythatitisstillnotcleartome.canyouexplainwithsomeexampleswhatisnotincludedinmysolutionandwhatyoumeanwithemptysubsets?btw:imisreadthatthesethasonly5elements.butithas6elements.thisisfixed.

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