What is the total number of combination of n different things taken 1,2,3`"…."`n at a time?

Options

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

Answer

To find the total number of combinations of \( n \) different things taken \( 1, 2, 3, \ldots, n \) at a time, we can follow these steps:

### Step 1: Understand the Combinations
The number of ways to choose \( r \) items from \( n \) items is given by the combination formula:
\[ C(n, r) = \frac{n!}{r!(n-r)!} \]
For our case, we need to consider combinations for \( r = 1, 2, 3, \ldots, n \).

### Step 2: Write the Total Combinations
The total number of combinations when taking \( 1, 2, 3, \ldots, n \) at a time is:
\[ C(n, 1) + C(n, 2) + C(n, 3) + \ldots + C(n, n) \]

### Step 3: Include \( C(n, 0) \)
To simplify the calculation, we can include \( C(n, 0) \) (which is equal to 1) in our sum:
\[ C(n, 0) + C(n, 1) + C(n, 2) + C(n, 3) + \ldots + C(n, n) \]
This gives us:
\[ 1 + C(n, 1) + C(n, 2) + C(n, 3) + \ldots + C(n, n) \]

### Step 4: Use the Binomial Theorem
According to the binomial theorem, we know that:
\[ (1 + 1)^n = C(n, 0) + C(n, 1) + C(n, 2) + \ldots + C(n, n) \]
This simplifies to:
\[ 2^n = C(n, 0) + C(n, 1) + C(n, 2) + \ldots + C(n, n) \]

### Step 5: Subtract \( C(n, 0) \)
Since \( C(n, 0) = 1 \), we can rewrite our total combinations as:
\[ C(n, 1) + C(n, 2) + C(n, 3) + \ldots + C(n, n) = 2^n - 1 \]

### Final Answer
Thus, the total number of combinations of \( n \) different things taken \( 1, 2, 3, \ldots, n \) at a time is:
\[ \text{Total combinations} = 2^n - 1 \]

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