Question

18. If 'A' and 'B' are any two non - empty sets then.
(i) n (AU B) = n(A) + n(B)
(ii) n(AUB) = n(A). n(B)
(iii) n(AUB) = n(A) + n(B) -- n(
A nB)
(iv) n(AUB) = n(A) - n(B) + n(A n B)
(A) only i
(B) i and ii
(C) i and iv
(D) only ii
​

Answer 1

Step-by-step explanation:

only 1 is correct answer

Answer 2

If A and B are any two non-empty sets then, n(A U B) = n(A) + n(B) - n(A∩B).

• n(A∪B) = n(A) + n(B) – n(A∩B) if A and B are two finite sets.
• Simply put, the range of factors withinside the union of units A and B is identical to the sum of the cardinal numbers of units A and B minus the intersection of these units.
• The figure below demonstrates this point.
• Furthermore, we might also additionally enlarge this to 3 units.
• If A, B, and C are distinct finite sets, then: n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)
• The Venn diagram suggests that the union of the 3 sets is the entire of the cardinal numbers of units A, B, and C, in addition to the not unusual place factors of the 3 units, omitting the common factors of units taken in pairs of.

Note:

The options to this question are incorrect. Only the third law of set is correct which is not part of the options.