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

To solve the problem of forming a team of three students from different sections without all members being from the same section, we first calculate the total ways to form a team and then subtract the cases where all members come from one section.

1. Calculate the total number of ways to choose 3 students from the 30 students available (8 from Section Alpha, 10 from Section Beta, and 12 from Section Gamma).
   The total number of ways is given by: \(C(30,3)\), where \(C(n,r)\) is the combination of \(n\) items taken \(r\) at a time. \(C(30,3) = \frac{30 \times 29 \times 28}{3 \times 2 \times 1} = 4060\)
2. Next, calculate the number of ways where all 3 students are from the same section. This involves summing up the combinations from each individual section:
   • Section Alpha: \(C(8,3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\)
   • Section Beta: \(C(10,3) = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\)
   • Section Gamma: \(C(12,3) = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220\)
3. Subtract the number of undesired combinations from the total combinations: \(4060 - 396 = 3664\).

Therefore, the number of ways to form a team where all members are not from the same section is 3664.