Question

Find no of sets A and B such that A n B = {3,5} and A U B = {2,3,5,7,8)

Answer 1

A $\cup$ B = {2,3,5,7,8}
A $\cap$ B = {3,5}

A and B both have only 3 and 5 common.
All the sets of A and B will have 3 and 5.
For the uncommon numbers, irrespective of the set they belong to , they will come in union and not in intersection.

2,7,8 are to be divided into 2 sets.
for each digit there are 2 choices : either belong to A or B but not in both.
 

2 * 2 * 2 = 8 such sets possible.

It's rather a Permutation Combination question

Answer 2

Given that : A∩B = {3,5} and A U B = {2,3,5,7,8}
Now we know that {3,5} are common to both sets A and B so they will be present in both sets A and B. as for rest of elements {2,7,8} for them to be a part of A U B they need to be exclusively be a part of A or B but not both so each of the elements have 2 choices i.e. an element belongs to A or an element belongs to B, There for there are 8 possible sets:
i) A={3,5} and B={2,3,5,7,8}   
ii) A={2,3,5} and B={3,5,7,8}   
iii) A={3,5,7} and B={2,3,5,8}   
iv) A={3,5,8} and B={2,3,5,7}   
v) A={2,3,5,7} and B={3,5,8}   
vi) A={2,3,5,8} and B={3,5,7} 
vii) i) A={3,5,7,8} and B={2,3,5}   
viii) A={2,3,5,7,8} and B={3,5}

Another way to come to conlusion is truth table method :

Truth Table for element {3,5}

Present in A	Present in B	Result
True	True	Possible case
True	False	Not possible for {3,5} need to be in both
False	True	Not possible for {3,5} need to be in both
False	False	Not possible for {3,5} need to be in both

 

Truth Table for element {2,7,8}

Present in A	Present in B	Result
True	True	Not possible, should be present in atmost one for union
True	False	Possible, present in A
False	True	Possible, present in B
False	False	Not possible, Should be in atleast 1 to be present in Union

Therefore we can conclude {3,5} we have 1 possibilty each and for {2,7,8} we have 2 possibility in each. So total possible combinations = 1*1*2*2*2 = 8 cases