Set Theory

Given data:
Let:
number of students who studied Mathematics,
number of students who studied Physics,
number of students who studied Chemistry.

Given conditions:

Number of students studying two subjects:

Number of students studying none:

Using the formula for the union of three sets:

Substituting the values:
where is the number of students who studied all three subjects.

Simplifying:

Finding the range for :
From the given bounds:
Therefore:
Calculating :

Using the principle of inclusion-exclusion for three sets , , and , we have:

Given:

Since , substitute the values and solve for :

Thus, the maximum number of students who passed in all three subjects is 10.

Given the equation:

Step 1. By trial, we find that and satisfy the equation, as:

Step 2. Thus, the only solution in is .

Step 3. Calculate :

The Correct Answer is: 46

The problem asks for the value of , where is the total number of functions satisfying the condition for all , with and being the least possible natural number.

Concept Used:

The solution involves combinatorial counting principles. A function is defined by assigning to each element exactly one element from . The total number of such functions is .

In this problem, the domain is and the codomain is the power set of A, . A special condition is imposed on the function, which restricts the possible outputs for each input. To find the total number of such functions, we need to determine the number of valid choices for the output for each input and then multiply these numbers together, as the choice for each input is independent.

Step-by-Step Solution:

Step 1: Analyze the domain and codomain of the function.

The domain is the set . The size of the domain is .

The codomain is the power set of A, , which is the set of all subsets of A. The size of the power set is .

A function maps each element to a subset of A, denoted by .

Step 2: Apply the given condition to determine the number of choices for each function value.

The condition is that for any element , we must have . This means that for each , the subset that the function assigns must contain the element itself.

Let's find the number of possible subsets for an arbitrary element, say . The subset must be a subset of and must contain .

A subset of is formed by deciding for each of the 7 elements of whether to include it or not. For the subset :

• The inclusion of the element is fixed (it must be in the set). There is only 1 choice for this element.
• For the other elements in , each can either be included in or not. This gives 2 choices for each of these 6 elements.

Therefore, the total number of valid subsets that can be assigned to is .

Step 3: Calculate the total number of such functions.

The choice of the image for each is independent. The total number of functions is the product of the number of choices for each element in the domain.

Since , and for each of the 7 elements there are possible images, the total number of functions is:

Step 4: Express the total number in the form with the least .

We are given that the total number of functions is , where and is the least possible value. We have:

For to be the least possible natural number (and ), we must choose the smallest possible base, which is the prime base of the number. In this case, the base is 2.

So, we let . Then, we have:

The values are and .

Final Computation & Result:

Step 5: Calculate the value of .

With and , the sum is:

The value of is 44.

Define symmetric relations: A relation is symmetric if . A relation is reflexive if for all .

Count total relations:

Count reflexive relations: Reflexive pairs: (4 pairs). Remaining symmetric pairs: (6 pairs).

Count symmetric relations:

Non-reflexive symmetric relations:

Thus, the answer is: 960

To solve this problem, we first identify the elements of the set . We need to find pairs such that they satisfy the equation with being natural numbers. Let's explore possible values of and determine corresponding values.

Step 1: Since and are natural numbers, they must be positive integers. Therefore, the minimum value for is 1.

Step 2: Calculate possible values. We find values of by manipulating the equation:

. is an integer if is divisible by 3.

Step 3: Check values of :

Step 4: The set consists of these pairs: . Thus, and .

Step 5: Set . Consequently, . Therefore, the cardinality of , .

Step 6: Calculate the number of one-one functions from to . A one-one (bijective) function exists when . The number of such functions is equal to the number of permutations of , which are represented as .

Conclusion: The number of one-one functions from to is 24, which lies within the given range of 24 to 24.

We are given the word GARDEN, which consists of the following letters: G, A, R, D, E, N. Among these letters, the vowels are A and E. To find the probability that the selected word will NOT have vowels in alphabetical order, we proceed as follows:
Step 1: Total number of arrangements.
Since there are 6 distinct letters in the word GARDEN, the total number of ways to arrange these letters is:
Step 2: Number of favorable cases (vowels in alphabetical order).
For the vowels A and E to be in alphabetical order, the positions of A and E must be such that A appears before E. The total number of ways to arrange the 6 letters such that A appears before E is:
Step 3: Probability calculation.
The probability that the selected word will have vowels in alphabetical order is: Therefore, the probability that the selected word will NOT have vowels in alphabetical order is:

• Partition with one subset (trivial partition):
• Partitions with two subsets:
• Partition with three subsets:

Step 1: Define the reflexive and transitive conditions. A relation is reflexive if it contains for all , meaning it must have . Since and are included, transitivity requires to be included.

Step 2: Count valid relations. The possible additional elements are and , which must be avoided to prevent symmetry.

The valid relations satisfying reflexivity and transitivity but not symmetry are counted, giving: Thus, the answer is .

Step 1: Understanding the set

The set consists of , the set of the first ten primes, and , the set of all possible products of distinct elements of . Thus, , since there are subsets of , excluding the empty subset.

Step 2: Counting the pairs where divides

For each , divides exactly half of the elements of , as for every product that doesn't contain , there is a corresponding one that does. Hence, for each , there are 512 elements in divisible by .

Step 3: Total number of pairs

Since there are 10 elements in , the total number of ordered pairs such that divides is:

Final Answer:

The correct answer is .