let set M={M1,M2,M3,...,M10},where all 10 elements are distinct.if sets are formed which contain 1 or more elements of set M,with subscript of each elements in a specific set being an integral multiple of the smallest subscript in the set,how many such sets can be formed?

• The question can be analysed base on the smallest subscript preesent in the set.
  
  Let a possible set has the smallest subscript as (M1). Now let us consider all the possibilities with the given condition
  Here M2 can be included ar excluded; M3 can be included or excluded; same is the case with M4,M5,....M10.
  Note that as the smallest subscript is M1, there is no need to certainly exclude any element, because every number is a multiple of 1
  
  So, for selecting M2, there are two possibilities(yes or No)same is the case with M3(yes or no).......same for M10(yes or no)
  
  Total possibilities with M1 as the lowest subscript is 2*2*2*....9times = 2^9= 512 possible sets
  
  Now lets consider the case of M2 as the lowest possible subscript, the definete elements that must be excluded from this set are M3,M5,M7,M9
  
  M4 can be included or excluded(two ways) same is the case with M6 andM8 and M10, so total possibilities is 2*2*2*2= 16 ways
  Similarly the number of possible sets with
  
  M3 as lowest element is 2*2=4; only M6 and M9 are considered
  M4 as the lowest element is 2; only M8 is considered
  M5 as the lowest element is 2; only M10 is considered
  M6 as the lowest element is only one set
  Only one set each with M7,M8,M9,M10 as lowest element
  
  So, total sets is 512(M1)+16(M2)+4(M3)+2(M4)+2(M5)+1(M6)+1(M7)+1(M8)+1(M9)+1(M10)
  
  Answer is 541 sets
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