Question

Out of 8 students in Section Alpha, 10 students in Section Beta and 12 students in Section Gamma, a teacher wants to create a team of three to represent the school in inter-school quiz competition. In how many ways the team can be formed such that all the members of the team are not from the same section ?

• A. 3664
• B. 396
• C. 4060
• D. 8120

Hint:

For problems with conditions like "not all from the same...", calculating the complementary event (all from the same) and subtracting it from the total unconstrained possibilities is usually much faster than calculating the constrained possibilities directly.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to form a team of 3 students from a total of \(8 + 10 + 12 = 30\) students. The condition is that the team must not consist of students who all belong to the same section. The easiest way to solve this is by using the complement principle: Total ways to form a team minus the ways to form a team where all members are from the same section.


Step 2: Key Formula or Approach:
Total ways to select \(r\) items from \(n\) items = \(\binom{n}{r}\)
Required ways = (Total ways to select 3 students) - (Ways to select 3 students from Alpha) - (Ways to select 3 students from Beta) - (Ways to select 3 students from Gamma)


Step 3: Detailed Explanation:
Total number of students = \(8 + 10 + 12 = 30\).
Total ways to select a team of 3 from 30 students: \[ \binom{30}{3} = \frac{30 \times 29 \times 28}{3 \times 2 \times 1} = 5 \times 29 \times 28 = 4060 \] Now, let's calculate the number of ways where all 3 members are from the same section:
Case 1: All 3 from Section Alpha (8 students): \[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \] Case 2: All 3 from Section Beta (10 students): \[ \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \] Case 3: All 3 from Section Gamma (12 students): \[ \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \] Total ways to form a team with all members from the same section = \(56 + 120 + 220 = 396\).
Finally, subtract this from the total ways: \[ \text{Required ways} = 4060 - 396 = 3664 \]

Step 4: Final Answer:
The team can be formed in 3664 ways.