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9 votes
9 votes
N(A – B) + n(A ∩ B ) =
n(A)
n(B)

n(AUB)

n(B−A)

User Jonathan Wakely
by
3.3k points

1 Answer

17 votes
17 votes

Answer:

n(A)

Explanation:

Refer to the attachment that I drew. First, draw a diagram of two sets A and B together. The diagram of A - B represents all elements in A only - the intersection part does not count in.

The diagram of A ∩ B represents the intersection part of A and B, meaning both A and B have same elements and that belong to the intersection part.

Now, add two shades together and it’ll result in set A being full-shade while set B only has the intersection part shaded but not full part. Let’s consider each choices:

  1. n(A) is correct as it represents the number of elements in set A, including the intersection part. From the diagram, this choice matches.
  2. n(B) is incorrect because the diagram has set A shaded, not set B.
  3. n(A∪B) is incorrect because A ∪ B means the union of both sets. The diagram of union set of A and B would be full-shade for both sets.
  4. n(B-A) is incorrect since the diagram of B - A will have only set B shaded excluding intersection part.

Therefore, the first choice n(A) is correct.

N(A – B) + n(A ∩ B ) = n(A) n(B) n(AUB) n(B−A)-example-1
User Aestrro
by
2.4k points