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