If some or all of n objects are taken at a time, then the number of combinations is `2^(n) - 1`.

Answer

To solve the problem, we need to find the sum of combinations of n objects taken at a time, specifically \( \binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \ldots + \binom{n}{n} \).

### Step-by-Step Solution:

1. **Understanding the Binomial Theorem**:
The Binomial Theorem states that:
\[ (A + B)^n = \sum_{k=0}^{n} \binom{n}{k} A^{n-k} B^k \]
For our case, we can set \( A = 1 \) and \( B = 1 \).

2. **Applying the Binomial Theorem**:
Substituting \( A \) and \( B \) into the Binomial Theorem gives us:
\[ (1 + 1)^n = \sum_{k=0}^{n} \binom{n}{k} 1^{n-k} 1^k \]
This simplifies to:
\[ 2^n = \sum_{k=0}^{n} \binom{n}{k} \]

3. **Identifying the Terms**:
The sum \( \sum_{k=0}^{n} \binom{n}{k} \) includes all combinations from 0 to n. We can express this as:
\[ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} = 2^n \]

4. **Separating the Terms**:
We know that \( \binom{n}{0} = 1 \). Therefore, we can rewrite the equation:
\[ 1 + \left( \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} \right) = 2^n \]

5. **Isolating the Desired Sum**:
To find the sum \( \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} \), we subtract 1 from both sides:
\[ \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} = 2^n - 1 \]

6. **Conclusion**:
Thus, we have shown that the sum of combinations of n objects taken at a time (from 1 to n) is:
\[ \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} = 2^n - 1 \]

### Final Result:
The number of combinations when taking some or all of n objects at a time is \( 2^n - 1 \).