here,nothing is given about referential integrity so,we apply some case

CASE 1 (default): All values of R1(R) and R2(R) are same then 20*10=200

CSAE 2: suppose R2(R) references R1(R) (foreign key) and all the values are same in R2(R) then,

1 tuple of R1(R) matches with 10 tuples of R2(R) then no of combination=10

and 19 tuples left in R1 which also come in resultant table with NULL values for R2 tables attributes value

so,19 no of more tuples

so, total =10+19=29

CASE 3: if we change RI make R1(R) references R2(R) then if all values R1(R) is same

then, also as in case 2, i.e. 20+9=29

CASE 4: if r be act as primary key (unique) for both the table then for EX

R1

P |
Q |
R |
S |

a |
d |
1 |
o |

b |
e |
2 |
u |

c |
f |
3 |
g |

R2

R |
G |
T |
U |

4 |
k |
n |
q |

5 |
l |
o |
r |

6 |
m |
p |
s |

R1(leftouter join)R2

P |
Q |
R |
S |
R |
G |
T |
U |

a |
d |
1 |
o |
n |
n |
n |
n |

b |
e |
2 |
u |
n |
n |
n |
n |

e |
f |
3 |
g |
n |
n |
n |
n |

so,all left part of table present with null values is right part of table

so,cardinality of left table will be maxm no of tuples so, 20 will be maxm