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**sid_178** a)Since cardinality of the set is 21 .There are 2^21 subsets in total for the given set .Such a collection of all the subsets of the set is also known as power set .

b)There are in total 7 numbers in the given set which are divisible by 7 namely{0,3,6,9,12,15,18}So there is only one way in which all these seven elements can be dealt with . Rest 14 elements have two ways in which they can be dealt with .Either they are present in the subset or not present in the subset .Therefore there are 2^14 subsets in total which contain all the numbers divisible by three of the given set .

__Alternatively__ this sum is 14C0 +14C1 +14C2 +.....14C14 =2^14 (BECAUSE ..you take either no element from the remaining 14elements or one or two and so on )

c)Similarly,7,9 and 15 are already there in the subsets we are left with 18 elements which can be dealt in two ways ..so answer becomes 2^18.

d)answer to this is total nu,mber of subsets -those subsets which contain 7,9and 15=2^21-2^18.

e)Answer to this is 2^16.Since elements {7,9,15} and {11 and 12 }can be dealt only in one way ..rest 16 elements in two ways ..either present or not present .