As an experienced tutor registered on UrbanPro, I'd approach this problem systematically. Firstly, let's denote:

• HH as the number of students who play hockey,
• BB as the number of students who play basketball,
• CC as the number of students who play cricket.

Given information:

• H=23H=23,
• B=15B=15,
• C=20C=20,
• H∩B=7H∩B=7,
• B∩C=5B∩C=5,
• H∩C=4H∩C=4,
• Total students=60Total students=60,
• Students not playing any game=15Students not playing any game=15.

Now, we can use the principle of inclusion-exclusion to find the answers.

(i) How many play hockey, basketball, and cricket? To find this, we'll use the formula: n(H∪B∪C)=n(H)+n(B)+n(C)−n(H∩B)−n(B∩C)−n(H∩C)+n(H∩B∩C)n(H∪B∪C)=n(H)+n(B)+n(C)−n(H∩B)−n(B∩C)−n(H∩C)+n(H∩B∩C)

n(H∪B∪C)=23+15+20−7−5−4+0=42n(H∪B∪C)=23+15+20−7−5−4+0=42

(ii) How many play hockey but not cricket? n(H∩C‾)=n(H)−n(H∩C)n(H∩C)=n(H)−n(H∩C) n(H∩C‾)=23−4=19n(H∩C)=23−4=19

(iii) How many play hockey and cricket but not basketball? n(H∩C∩B‾)=n(H∩C)−n(H∩B∩C)n(H∩C∩B)=n(H∩C)−n(H∩B∩C) n(H∩C∩B‾)=4−0=4n(H∩C∩B)=4−0=4

So, as an experienced tutor on UrbanPro, I'd explain these concepts to my students using clear and concise language, ensuring they understand the logic behind the calculations.

As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into the properties of sets to prove the given statement.

To prove: A−(A∩B)=A−BA−(A∩B)=A−B

Let's break it down step by step using set operations:

1. Left-hand side (LHS) of the equation: A−(A∩B)A−(A∩B)

   • This means we're taking all elements in set A that are not in the intersection of sets A and B.

2. Right-hand side (RHS) of the equation: A−BA−B

   • This means we're taking all elements in set A that are not in set B.

To prove the equality, we need to show that the LHS is equal to the RHS.

Proof:

1. Let's start with the LHS: A−(A∩B)A−(A∩B)

   • This means we're removing from set A all elements that are in both sets A and B.

2. Now, let's consider the RHS: A−BA−B

   • This means we're removing from set A all elements that are in set B.

3. If we compare the two operations:

   • On the LHS, we're removing elements from A that are in both A and B.
   • On the RHS, we're removing elements from A that are in set B.

4. It follows that whatever elements were in both A and B, they are removed in both operations.

   • So, A−(A∩B)A−(A∩B) and A−BA−B both remove the same elements from set A.

5. Therefore, A−(A∩B)=A−BA−(A∩B)=A−B, proving the equality.

This completes the proof, demonstrating that for all sets A and B, A−(A∩B)A−(A∩B) is indeed equal to A−BA−B. This principle is fundamental in set theory and can be useful in various mathematical and logical contexts.

As a seasoned tutor registered on UrbanPro, I'm glad to assist you with this question. UrbanPro is indeed a fantastic platform for online coaching and tuition, providing a conducive environment for both tutors and students to engage in effective learning experiences.

Let's tackle the given sets and operations:

Given:

• U = {1, 2, 3, 4, 5, 6, 7}
• A = {2, 4, 6}
• B = {3, 5}
• C = {1, 2, 4, 7}

(i) We need to find A′ ∪ (B ∩ C′)

To find A′ (complement of A), we need to consider all the elements in U that are not in A: A′ = {1, 3, 5, 7}

Next, let's find C′ (complement of C), which includes all elements in U that are not in C: C′ = {3, 5, 6}

Now, we need to find the intersection of B and C′: B ∩ C′ = {5}

Finally, we take the union of A′ and (B ∩ C′): A′ ∪ (B ∩ C′) = {1, 3, 5, 7} ∪ {5} = {1, 3, 5, 7}

(ii) We need to find (B – A) ∪ (A – C)

First, let's find (B – A), which means elements in B that are not in A: B – A = {3, 5}

Next, let's find (A – C), which means elements in A that are not in C: A – C = {6}

Now, we take the union of (B – A) and (A – C): (B – A) ∪ (A – C) = {3, 5} ∪ {6} = {3, 5, 6}

So, the solutions are: (i) A′ ∪ (B ∩ C′) = {1, 3, 5, 7} (ii) (B – A) ∪ (A – C) = {3, 5, 6}

If you have any further questions or need clarification, feel free to ask!

As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best online coaching tuition platforms available for students seeking academic assistance. Now, let's tackle your set theory problem using the given sets U, A, and B.

Given: U = {1, 2, 3, 4, 5, 6} A = {2, 3} B = {3, 4, 5}

First, let's find the complements of sets A and B: A′ (complement of A) = {1, 4, 5, 6} (Elements in U but not in A) B′ (complement of B) = {1, 2, 6} (Elements in U but not in B)

Now, let's find the intersection of the complements of A and B: A′ ∩ B′ = {1, 6} (Elements common to both A′ and B′)

Next, let's find the union of sets A and B: A ∪ B = {2, 3, 4, 5} (All elements in both A and B without repetition)

Finally, let's find the complement of the union of sets A and B: (A ∪ B)′ (complement of A ∪ B) = {1, 6} (Elements in U but not in A ∪ B)

We've shown that A′ ∩ B′ = (A ∪ B)′, which demonstrates De Morgan's Law for sets.