Number of ways of forming a team consisting of at least 2 and atmost (n - 2) students from n students is

Options

A. `2^(n)2n`
B. `2^(n)2n-2`
C. `2^(n)2n-4`
D. `2^(n)-2n-1`

Answer

To solve the problem of finding the number of ways to form a team consisting of at least 2 and at most (n - 2) students from n students, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to form teams with at least 2 students and at most (n - 2) students from a total of n students.

2. **Identifying the Range**:
The number of students in the team can vary from 2 to (n - 2). Therefore, the possible sizes of the team are:
- 2 students
- 3 students
- 4 students
- ...
- (n - 2) students

3. **Using Combinations**:
The number of ways to choose k students from n students is given by the combination formula \( C(n, k) \) or \( \binom{n}{k} \). Thus, we can express the total number of ways to form the teams as:
\[ C(n, 2) + C(n, 3) + C(n, 4) + \ldots + C(n, n-2) \]

4. **Using the Binomial Theorem**:
According to the binomial theorem, the sum of combinations can be expressed as:
\[ \sum_{k=0}^{n} C(n, k) = 2^n \]
We can use this to find our required sum by subtracting the combinations we do not want (i.e., \( C(n, 0) \), \( C(n, 1) \), \( C(n, n-1) \), and \( C(n, n) \)):
\[ C(n, 2) + C(n, 3) + \ldots + C(n, n-2) = \sum_{k=0}^{n} C(n, k) - C(n, 0) - C(n, 1) - C(n, n-1) - C(n, n) \]

5. **Calculating the Unwanted Combinations**:
The unwanted combinations are:
- \( C(n, 0) = 1 \)
- \( C(n, 1) = n \)
- \( C(n, n-1) = n \)
- \( C(n, n) = 1 \)
Therefore, the total of unwanted combinations is:
\[ 1 + n + n + 1 = 2n + 2 \]

6. **Final Calculation**:
Now, substituting back into our equation:
\[ C(n, 2) + C(n, 3) + \ldots + C(n, n-2) = 2^n - (2n + 2) \]
Simplifying this gives:
\[ C(n, 2) + C(n, 3) + \ldots + C(n, n-2) = 2^n - 2n - 2 \]

### Final Answer:
The number of ways of forming a team consisting of at least 2 and at most (n - 2) students from n students is:
\[ 2^n - 2n - 2 \]