 Hello and welcome to the session. In this session, we will discuss the following question and the question says, from the given REN diagram, find the following sets. First part is A, second part is B, third part is A union B, fourth part is A intersection B. Also verify that number of elements in A union B is equal to number of elements in A plus number of elements in B minus number of elements in A intersection B. Let's start the solution now. We are given this REN diagram in which the two intersecting circles A and B represent the two overlapping sets A and B respectively. The elements of the set A are written inside this circle A and the elements of the set B are written inside this circle B. In the first part, we have to find the set A. We know that set A contains all the elements which are inside the circle A. So, the set A contains all the elements which are inside this shaded portion. So, A is equal to the set containing the elements 1, 3, 5, 6, 7, 9. This is the answer for the first part. Now in the second part, we have to find the set B. Now the set B contains all the elements which are inside the circle B. So, set B contains all the elements which are in this shaded portion. That is, B is equal to the set containing the elements 2, 3, 4, 6, 9, 10, 12. This is our answer for the second part. In the third part, we have to find A union B. Now A union B contains elements which are in set A or set B or both. So, A union B contains elements which are in this shaded portion. So, A union B is equal to the set containing the elements 1, 2, 3, 4, 5, 6, 7, 9, 10, 12. This is our answer for the third part. In the fourth part, we have to find A intersection B. Now A intersection B contains elements which are in set A and in set B. That is, A intersection B contains elements which are common in both the sets A and B. Now the common elements between the two sets is written in this common portion between the two circles A and B. So, A intersection B contains elements which are in this shaded portion. Thus, A intersection B is equal to the set containing the elements 3, 6, 9. Now we have to verify that number of elements in A union B is equal to number of elements in A plus number of elements in B minus number of elements in A intersection B. Let us see the right-hand side of the given equality first. So, the right-hand side is number of elements in A plus number of elements in B minus number of elements in A intersection B. In part 1, we found A. So, A contains 6 elements. So, number of elements in A is equal to 6. Also, in part 2, we found B and we can see that the set B contains 7 elements. So, the number of elements in B is equal to 7. From part 3, we can see that A union B contains 10 elements. So, number of elements in A union B is equal to 10. Similarly, from part 4, we can say that number of elements in A intersection B is equal to 3. So, the right-hand side of the equality becomes number of elements in A which is equal to 6 plus number of elements in B which is equal to 7 minus number of elements in A intersection B which is equal to 3. This is equal to 13 minus 3 which is equal to 10. Now, let us see the left-hand side of the given equality which is equal to number of elements in A union B. This is equal to 10. So, right-hand side is equal to left-hand side. Thus, we have proved that number of elements in A union B is equal to number of elements in A plus number of elements in B minus number of elements in A intersection B. With this we end our session. Hope you enjoyed the session.