Question

if n(A)=38, n(B)=20 and n(A intersects B)= 12, then find n(A U B) ; find n(B- A) ; find n(A-B)​

Answer 1

To Find :-

• we need to find the Value of n(A U B)
• n(B - A)
• n(A - B)

Given :

• n(A) = 38
• n(B) = 20
• n(A ∩ B) = 12

By using Formula :-

• ◼ n( A U B ) + n(A ∩ B) = n(A) + n(B)

⇛n(A U B) + 12 = 38 + 20

⇛n(A U B) = 58 - 12

⇛n(A U B) = 46

Now,

• n(A - B)

⇛ n(A - B) = n(A) - n(A ∩ B)

⇛ n(A - B) = 38 - 12

⇛ n(A - B) = 26

Now,

• n(B - A)

⇛ n(B - A) = n(B) - n(A ∩ B)

⇛ n(B - A) = 20 - 12

⇛ n(B - A) = 8

Hence,

• n(A U B) = 46
• n(A - B) = 26
• n(B - A) = 8

━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

Answer:

Step-by-step explanation:

n(AUB) means the the number of elements in both A & B.

And that includes the ones in common too.

n(A∩B) means the ones that in common.

n(A-B) refers to the elements that belong to A alone.

That means the elements in A excluding the common elements.

n(B) is the number of elements that belong to B and that includes the common ones too.

Now coming to your question.

The total number of elements that can fit in both the sets is

$36.$

While the number of common elements is$16.$

And also given that n$(A-B)=15.$

Now according to the definitions above....n(A) should include the ones that belong exclusively to A alone and also the ones in common.

So,

$n(A)=n(A-B)+n(A∩B)$

= $15+16$

= $31$

Therefore, n(A) is  $31.$

Now when n(AUB) is $36$and n(A) is $31,$ it is clear that the elements exclusive to B alone is 5, i$.e.36-31.$

Now$n(B)=n(A∩B)+n(B-A)$

Therefore,$n(B) is 16+5 = 21.$

$n(B)=21.$

hope it helps

:)