What is the formula for n(AUBUCUD)
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Answered by
24
Hey Friend,
This is the formula for The above:
Note ∑ ---> stands for Summation
for e.g., ∑n(A) = |A| + |B| + |C| + |D|
∑n|A| - ∑n|A∩B| + ∑n|A∩B∩C| - n|A+B+C+D|
This is the formula for The above:
Note ∑ ---> stands for Summation
for e.g., ∑n(A) = |A| + |B| + |C| + |D|
∑n|A| - ∑n|A∩B| + ∑n|A∩B∩C| - n|A+B+C+D|
Answered by
4
Answer:
→ n(A∪B∪C∪D) = n(A) + n(B) + n(C) + n(D) - n(A∩B) - n(B∩C) - n(C∩D) - n(A∩C) - n(B∩D) - n( A∩D) + n(A∩B∩C∩D)
Step-by-step explanation:
Let A and B be two intersecting subsets of U.
In counting the elements of (AUB), the elements of A∩B are counted twice , once in counting the elements of a and second time in counting the elements of 8.
→n(A∪B) = n(A) + n(B) -n(A∩B)
from the Venn Diagram,
(i) n(A-B) + n(A∩B) = n(A)
(ii) n(B-A) + n(A∩B) = n(B)
(iii) n(A-B) + n(A∩B) + n(B-A) = n(A∪B).
→ n(A∪B∪C) = n(A) + n(B) +n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C)
→ n(A∪B∪C∪D) = n(A) + n(B) + n(C) + n(D) - n(A∩B) - n(B∩C) - n(C∩D) - n(A∩C) - n(B∩D) - n( A∩D) + n(A∩B∩C∩D)
THANKS.
#SPJ3.
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