Each student in a class of 40 studies at least one of the subjects English, Maths, and Eco; 16 study English, 22 Maths, and 26 Eco, 5 study English and Eco, 14 Maths and Eco, and 2... Each student in a class of 40 studies at least one of the subjects English, Maths, and Eco; 16 study English, 22 Maths, and 26 Eco, 5 study English and Eco, 14 Maths and Eco, and 2 study English, Maths, and Eco. Find the number of students who study (i) English and Maths; (ii) English, Maths but not Eco. Read more

Steps to Solve

1. Identifying the variables Let:

• $E$: Number of students studying English
• $M$: Number of students studying Maths
• $C$: Number of students studying Eco
• $n(E)$: Students studying English = 16
• $n(M)$: Students studying Maths = 22
• $n(C)$: Students studying Eco = 26
• $n(E \cap C)$: Students studying both English and Eco = 5
• $n(M \cap C)$: Students studying both Maths and Eco = 14
• $n(E \cap M \cap C)$: Students studying all three = 2

2. Applying the principle of inclusion-exclusion Use the formula for the total number of students studying at least one subject: $$ n(E \cup M \cup C) = n(E) + n(M) + n(C) - n(E \cap M) - n(E \cap C) - n(M \cap C) + n(E \cap M \cap C) $$ Substituting in the known values, we can solve for $n(E \cap M)$.

3. Setting up the equation We can rearrange the inclusion-exclusion principle to isolate $n(E \cap M)$: $$ 40 = 16 + 22 + 26 - n(E \cap M) - 5 - 14 + 2 $$ Simplifying gives: $$ 40 = 47 - n(E \cap M) $$

4. Solving for $n(E \cap M)$ Rearranging the equation leads to: $$ n(E \cap M) = 47 - 40 = 7 $$

5. Finding the number of students studying English and Maths but not Eco To calculate this, use: $$ n(E \cap M \setminus C) = n(E \cap M) - n(E \cap M \cap C) $$ Substituting the values we found, $$ n(E \cap M \setminus C) = 7 - 2 = 5 $$