A) 4000
due to the restrictions imposed by the functional dependencies we can vary values like:
keep B and C value as constant and loop through all values of A, this gives us 20 possible combinations for the value set to appear in the Relation R.
next on getting exhausted with the above situation we try to change the value of B, keeping C as before but FDs stops us from doing that and the FD set asks us to change the value for C also, therefore for possible combinations of BC we get 10 different sets for which each element of set is associated with 20 values of A. 10 different because we simultaneously change value for C as we change for B as we need maximum possible tuples in the result when performing the Self Join on C.
On Self join we see that for such a story the same values of C are to be combined with the same values of C only Because the join is on C.
for instance
A |
B |
C |
---|
$a_1$ |
$b_1$ |
$c_1$ |
$a_2$ |
$b_1$ |
$c_1$ |
$a_1$ |
$b_2$ |
$c_2$ |
$a_2$ |
$b_2$ |
$c_2$ |
for first row in the Relation we have with same C 20 different combinations, similarly for the second row on same value of C there that is $c_1$
then same is true for the third row onwards which has 20 different combinations on same value of C i.e. $c_2$ and so on for such 10 possible $b_i$ and $c_i$
Hence, (20*20) * 10 = 4000