Start with a grid of newspapers (A-D) to families (1-5)

|1|2|3|4|5
-----------
A| | | | | |
B| | | | | |
C| | | | | |
D| | | | | |

**we know one of the families subscribes to all

|1|2|3|4|5
-----------
A|x| | | | |
B|x| | | | |
C|x| | | | |
D|x| | | | |

so done with Family 1 and also done with newspaper A since only 1 subscriber

** we also know that 2 families subscribe to B, we already have 1 so add another

|1|2|3|4|5
-----------
A|x| | | | |
B|x|x| | | |
C|x| | | | |
D|x| | | | |

so done with newspaper B

** we know that newspaper D has all 5 families subscribed

|1|2|3|4|5
-----------
A|x| | | | |
B|x|x| | | |
C|x| | | | |
D|x|x|x|x|x|

so done with newspaper D

At this point we are done with newspaper A, B, and D and families 1

The only way we can meet the last 2 criteria of:
1) 3 families subscribed to 2 newspapers
2) newspaper C has 3 subscribers is as follows:

|1|2|3|4|5
-----------
A|x| | | | |
B|x|x| | | |
C|x| |x|x| |
D|x|x|x|x|x|

so person 5 is only subscribed to 1 newspaper in the end


Hope this helps