In a class of $200$ students numbered $1$ to $200$, those whose number is divisible by $2$ opted for Literature, those whose number is divisible by $3$ opted for History, and those whose number is divisible by $7$ opted for Philosophy. Then the number of students who did not opt for any of the three courses is:
58
Understanding the Problem: Students and Course Choices
We have a class of 200 students, numbered from 1 to 200. Different groups of students chose specific subjects based on their number being divisible by certain integers:
- Students whose number is divisible by 2 opted for Literature.
- Students whose number is divisible by 3 opted for History.
- Students whose number is divisible by 7 opted for Philosophy.
The goal is to find the number of students who did not opt for any of these three subjects (Literature, History, or Philosophy).
Defining Sets for Course Enrollment
To solve this, we can use sets and the Principle of Inclusion-Exclusion. Let:
- $S$ be the set of all students in the class. $|S| = 200$.
- $L$ be the set of students who opted for Literature (number divisible by 2).
- $H$ be the set of students who opted for History (number divisible by 3).
- $P$ be the set of students who opted for Philosophy (number divisible by 7).
We want to find the number of students outside the union of these sets, which is $|S| - |L \cup H \cup P|$.
Calculating the Number of Students for Each Subject
First, let's find the count for each set:
| Subject | Condition (Divisible by) | Calculation | Number of Students |
|---|---|---|---|
| Literature (L) | 2 | $\lfloor \frac{200}{2} \rfloor$ | 100 |
| History (H) | 3 | $\lfloor \frac{200}{3} \rfloor$ | 66 |
| Philosophy (P) | 7 | $\lfloor \frac{200}{7} \rfloor$ | 28 |
Calculating Students in Combined Subjects
Next, we find the number of students who opted for combinations of subjects. This involves finding numbers divisible by the least common multiple (LCM) of the subject divisors.
| Subjects Combined | Condition (Divisible by LCM) | Calculation | Number of Students |
|---|---|---|---|
| Literature and History (L ∩ H) | LCM(2, 3) = 6 | $\lfloor \frac{200}{6} \rfloor$ | 33 |
| Literature and Philosophy (L ∩ P) | LCM(2, 7) = 14 | $\lfloor \frac{200}{14} \rfloor$ | 14 |
| History and Philosophy (H ∩ P) | LCM(3, 7) = 21 | $\lfloor \frac{200}{21} \rfloor$ | 9 |
| Literature, History, and Philosophy (L ∩ H ∩ P) | LCM(2, 3, 7) = 42 | $\lfloor \frac{200}{42} \rfloor$ | 4 |
Applying the Principle of Inclusion-Exclusion
The Principle of Inclusion-Exclusion helps us find the total number of students who opted for at least one subject ($|L \cup H \cup P|$). The formula is:
$$ |L \cup H \cup P| = |L| + |H| + |P| - (|L \cap H| + |L \cap P| + |H \cap P|) + |L \cap H \cap P| $$
Substituting the values we calculated:
$$ |L \cup H \cup P| = 100 + 66 + 28 - (33 + 14 + 9) + 4 $$
First, sum the number of students in individual subjects:
$$ 100 + 66 + 28 = 194 $$
Next, sum the number of students in pairwise combinations:
$$ 33 + 14 + 9 = 56 $$
Now, apply the formula:
$$ |L \cup H \cup P| = 194 - 56 + 4 $$
$$ |L \cup H \cup P| = 138 + 4 $$
$$ |L \cup H \cup P| = 142 $$
So, 142 students opted for at least one of the three subjects.
Calculating Students Not Opting for Any Course
To find the number of students who did not opt for any subject, we subtract the number of students who opted for at least one subject from the total number of students:
Number of students not opting = Total students - $|L \cup H \cup P|$
Number of students not opting = $200 - 142$
Number of students not opting = $58$
Final Answer
Therefore, the number of students who did not opt for Literature, History, or Philosophy is 58.
Important Questions from Sets
The Cartesian product A × A has 16 elements among which are (0, 2) and (1, 3). Which of the following statements is/are correct?
1. It is possible to determine set A.
2. A × A contains the element (3, 2).
Select the correct answer using the code given below:
Consider the proper subsets of {1, 2, 3, 4}. How many of these proper subsets are a superset of the set {3}?
Let $f(x) = |x - 2| + |x - 8|$; $x \in R$. Then the set of all values of $x$, at which the function, $g(x) = f(f(x))$ is not differentiable, is:
If A = {x : x is a multiple of 7},
B = {x : x is a multiple of 5} and
C = {x : x is a multiple of 35}
Then which of the following is null set?
Consider the following statements in respect of two non-empty sets A and B :
1. A ∪ B = A ∩ B if A = B
2. A Δ B = ϕ if A = B
Which of the above statements is/are correct ?
