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1. The union of three pairwise disjoint sets containing 10, 15, and 25 elements will contain 50 elements.

   In the question, we are asked for the number of elements in the union of three pairwise disjoint sets if the sets contain 10, 15, and 25 elements.

   We assume the three sets to be A, B, and C, respectively.

   Thus, we are given that n(A) = 10, n(B) = 15, and n(C) = 25.

   Given that the three sets are pairwise disjoint, we know that the intersection of sets will be a  null set, that is,

   A ∩ B = B ∩ C = C ∩ A = A ∩ B ∩ C = {}.

   Thus, we can write that,

   n(A ∩ B) = 0,

   n(B ∩ C) = 0,

   n(C ∩ A) = 0, and

   n(A ∩ B ∩ C) = 0.

   Now, for the union of three sets, we need to know the formula:

   n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A  ∩ B ∩ C).

   Substituting the values, we get:

   n(A ∪ B ∪ C) = 10 + 15 + 25 – 0 – 0 – 0 + 0,

   or, n(A ∪ B ∪ C) = 50.

   Thus, the union of three pairwise disjoint sets containing 10, 15, and 25 elements will contain 50 elements.

   Learn more about sets at

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2. Answer:

     50 elements are in the union of three pairwise disjoint sets if the sets contain 10, 15, and 25 elements

   Step-by-step explanation:

   Let A, B and C be the three sets. We are given that the sets are pairwise disjoint. Therefore, we can write :

   n(A)=10

   n(B)=15

   n(C)=25

   n(A∩B)=0

   n(B∩C)=0

   n(C∩A)=0

   n(A∩B∩C)=0

   Number of elements in the union of these sets will be :

   n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C)

   Putting all the given values in the formula above, we get

   n(A∪B∪C)=10+15+25−0−0−0+0

   ⇒n(A∪B∪C)=50

   What is disjoint set?

   Two sets are said to be disjoint if they have no element in common. Equivalently, disjoint sets are sets whose intersection is the empty set.

   Learn more about disjoint sets here: https://brainly.ph/question/39533

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