500 students are taking one or more courses out of Chemistry, Physics, and Mathematics. Registration records indicate course enrolment as follows: Chemistry (329), Physics (186), Mathematics (295), Chemistry and Physics (83), Chemistry and Mathematics (217), and Physics and Mathematics (63). How many students are taking all 3 subjects?
53
This problem involves calculating the number of students taking all three subjects (Chemistry, Physics, and Mathematics) using the concept of set theory, specifically the Principle of Inclusion-Exclusion. We are given the total number of students taking at least one course and the number of students enrolled in individual subjects and pairs of subjects.
Students Enrollment Overview
We are provided with the following data regarding student enrollment in various courses:
| Category | Number of Students |
|---|---|
| Total students taking one or more courses (Chemistry \((\text{C})\), Physics \((\text{P})\), or Mathematics \((\text{M})\)) | \(|\text{C} \cup \text{P} \cup \text{M}| = 500\) |
| Students taking Chemistry | \(|\text{C}| = 329\) |
| Students taking Physics | \(|\text{P}| = 186\) |
| Students taking Mathematics | \(|\text{M}| = 295\) |
| Students taking Chemistry and Physics | \(|\text{C} \cap \text{P}| = 83\) |
| Students taking Chemistry and Mathematics | \(|\text{C} \cap \text{M}| = 217\) |
| Students taking Physics and Mathematics | \(|\text{P} \cap \text{M}| = 63\) |
Our goal is to find the number of students who are taking all 3 subjects, which means we need to find \(|\text{C} \cap \text{P} \cap \text{M}|\).
Inclusion-Exclusion Principle Application
The Principle of Inclusion-Exclusion is a powerful tool in combinatorics used to count the number of elements in the union of multiple sets. For three sets A, B, and C, the formula is:
\[|\text{A} \cup \text{B} \cup \text{C}| = |\text{A}| + |\text{B}| + |\text{C}| - (|\text{A} \cap \text{B}| + |\text{A} \cap \text{C}| + |\text{B} \cap \text{C}|) + |\text{A} \cap \text{B} \cap \text{C}|\]
In our problem, A represents Chemistry (C), B represents Physics (P), and C represents Mathematics (M). So, the formula becomes:
\[|\text{C} \cup \text{P} \cup \text{M}| = |\text{C}| + |\text{P}| + |\text{M}| - (|\text{C} \cap \text{P}| + |\text{C} \cap \text{M}| + |\text{P} \cap \text{M}|) + |\text{C} \cap \text{P} \cap \text{M}|\]
Calculating Students Taking All Subjects
Now, let's substitute the given values into the formula:
We know that \(|\text{C} \cup \text{P} \cup \text{M}| = 500\).
First, calculate the sum of individual enrollments:
- \(|\text{C}| + |\text{P}| + |\text{M}| = 329 + 186 + 295 = 810\)
Next, calculate the sum of enrollments in pairs of subjects:
- \(|\text{C} \cap \text{P}| + |\text{C} \cap \text{M}| + |\text{P} \cap \text{M}| = 83 + 217 + 63 = 363\)
Substitute these calculated sums into the Inclusion-Exclusion Principle formula:
\[500 = 810 - 363 + |\text{C} \cap \text{P} \cap \text{M}|\]
Perform the subtraction on the right side of the equation:
\[500 = 447 + |\text{C} \cap \text{P} \cap \text{M}|\]
To find the number of students taking all 3 subjects, rearrange the equation to isolate \(|\text{C} \cap \text{P} \cap \text{M}|\):
\[|\text{C} \cap \text{P} \cap \text{M}| = 500 - 447\]
\[|\text{C} \cap \text{P} \cap \text{M}| = 53\]
Therefore, 53 students are taking all three subjects: Chemistry, Physics, and Mathematics.
Important Questions from Numerical Computation
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A deposit in a bank, which pays interest on its deposits compounded daily, grows to Rs. 80,000 for 500 days and to 88,000 for 1000 days. What would be its value (in Rs.) for 1500 days?
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