Q.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?

Concept: Counting Principle — treat the 3 particular students as a single block when they all join, then handle the two disjoint cases.

Step 1 – Case 1: All 3 join.

If all 3 are in the party, we need $10-3 = 7$ more students from the remaining $25-3 = 22$ students.

Number of ways: $\binom{22}{7}$.

Step 2 – Case 2: None of the 3 join.

Then we choose all 10 from the other 22 students.

Number of ways: $\binom{22}{10}$. …

When a problem says "either all of them will join or none of them will join," it’s a classic signal to use the block method — a direct application of the Fundamental Counting Principle. The three students are inseparable in the sense that they act as one unit when they choose to go. But they also have the freedom to stay home together. So we have two completely separate scenarios, and we add the counts because they are mutually exclusive.

Let’s break it down.

1. Case 1: All three join.

   If all three are in the party, we have already chosen 3 out of the required 10. The remaining $10 - 3 = 7$ spots must be filled from the other $25 - 3 = 22$ students.

   The number of ways to choose these 7 is simply $\binom{22}{7}$.

2. Case 2: None of the three join.

   Here, the three students are completely out. So we need to choose all 10 members from the remaining 22 students.

   The number of ways is $\binom{22}{10}$.

3. Total ways.

   Since the two cases cannot happen at the same time (the three students either all go or all stay), we add the counts:

4. Compute the values. $\binom{22}{7} = \frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 170544.$ …