From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?
The excursion party can be chosen in 817190 ways.
Explanation
This is a combination problem with a constraint. The 3 specific students must either all join or none join, creating two mutually exclusive cases. We use the addition principle to sum the number of ways for both cases. The context shows similar problems where selections are made from groups using combinations (ⁿCᵣ notation) and results are added when cases are mutually exclusive.
Solution Steps
Step 1: Identify the two cases based on the constraint. The 3 students must either all join or none join. These are mutually exclusive cases.
Step 2: Case 1 - All 3 students join: If all 3 join, we need to choose remaining 7 students from the other 22 students (25 - 3 = 22). Number of ways = ²²C₇ = 22!/(7! × 15!) = 170544.
Step 3: Case 2 - None of the 3 students join: If none join, we need to choose all 10 students from the remaining 22 students. Number of ways = ²²C₁₀ = 22!/(10! × 12!) = 646646.
Step 4: Calculate total ways: By addition principle, Total = ²²C₇ + ²²C₁₀ = 170544 + 646646 = 817190 ways.