Er Diagram

Question

Consider the following entity relation diagram

If A has 50 entities,B has 20 entities and C has 200 entities. what is the maximum number of triples of entities that could be in the relationship Set R ?

(A) 50

(B) 100

(C) 1000

(D) 10,000

ans is A,C(given)

please explain how??

Comments

Vaishali06 according to me your answer and explanation is correct.

---

from korth

---

When we say AB->C ,then we should even make sure that A and B satisfy many to one relationship rght. .

But we took all possible combinations so How's that possible ?

Answer 1

Relation between A and B = R(A,B) = Many to One.

Relation between A and C = R(A,C) = Many to One.

Relation between B and C = R(B,C) = One to One.

Foreign Key of B and C will be added in A to mantain A's Relationship with B and C .

So now , A will be the Deciding Factor. 

And Total Tuples in the Relation R will be total Tuples Carried by A.

So, here you can easily decide the Correct one as , Option A contains a=50=t. And that is TRUE.

Hope This will Clear Your Doubt !!!

Comments

How answer is 50, 1000

If answer is 1000 then 50 is not maximum

Question is about maximum no. Of triples

So, only 1 answer is possible...

---

but 50 will be distinct unique values possible for table A  that would be minimum possible value of relationship set na

Answer 2

i dont know how it is 1000 but, i got 50 as the answer.
there is 1 to 1 relation between B and C,many-one between A and B, many to one between A and C.
B C pairs can be only 20 as there is one-one relation between them.
the requred triplet is <A B C>
there is many to one relation between A and <BC>
one element of A shud be related to atmost one pair of <BC>
there are 50 distinct values of A. each A can relate to any one of 20 <BC> combinations. so, maximum of 50 total number of tuples can be present

Comments

Minimum can be 50 tuples and maximum tuples  since 50×20 =1000 in relationship set

---

Kaluti then why not 10,000 ?

---

how maximum tuples  can be 10000 it will be 1000 na