No. of combinations by selecting one or more than one object out of n objects is:

Options

A. `2^(n-1)`
B. `2^(n)`-1
C. `2^(n)`
D. None of these

Answer

To find the number of combinations by selecting one or more than one object out of \( n \) objects, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to select at least one object from \( n \) objects. The selections can be one object, two objects, up to \( n \) objects.

2. **Using Combinations**:
The number of ways to choose \( k \) objects from \( n \) objects is given by the combination formula \( \binom{n}{k} \). Therefore, the total number of ways to choose one or more objects can be expressed as:
\[ \binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \ldots + \binom{n}{n} \]

3. **Applying the Binomial Theorem**:
According to the binomial theorem, we know that:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
If we set \( x = 1 \), we get:
\[ (1 + 1)^n = \sum_{k=0}^{n} \binom{n}{k} = 2^n \]

4. **Excluding the Empty Set**:
The sum \( \sum_{k=0}^{n} \binom{n}{k} \) includes the term \( \binom{n}{0} \), which represents the empty set (selecting no objects). Since we want to select one or more objects, we need to subtract this term:
\[ \sum_{k=1}^{n} \binom{n}{k} = 2^n - \binom{n}{0} = 2^n - 1 \]

5. **Final Result**:
Therefore, the total number of combinations by selecting one or more than one object out of \( n \) objects is:
\[ 2^n - 1 \]

### Conclusion:
The number of combinations by selecting one or more than one object out of \( n \) objects is \( 2^n - 1 \).