The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is

Options

A. `2^(20) - 1`
B. `2^(21)`
C. `2^(20)`
D. `2^(20) + 1`

Explanation

To solve the problem of choosing 10 objects from a total of 31 objects, where 10 are identical and 21 are distinct, we can break down the solution into several steps.

### Step 1: Understand the Composition of Objects
We have:
- 10 identical objects (let's call them A)
- 21 distinct objects (let's call them B1, B2, B3, ..., B21)

### Step 2: Determine the Number of Identical Objects to Choose
Since we have 10 identical objects, we can choose anywhere from 0 to 10 of these identical objects. Let's denote the number of identical objects chosen as \( x \). Thus, \( x \) can take values from 0 to 10.

### Step 3: Calculate the Remaining Objects to Choose
If we choose \( x \) identical objects, we need to choose \( 10 - x \) objects from the 21 distinct objects. The number of ways to choose \( 10 - x \) objects from 21 distinct objects is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of distinct objects and \( r \) is the number of objects to choose.

### Step 4: Set Up the Total Count
The total number of ways to choose the objects can be expressed as:
\[ \text{Total Ways} = \sum_{x=0}^{10} \binom{21}{10 - x} \]
This means we will sum the combinations for each possible value of \( x \) from 0 to 10.

### Step 5: Calculate Each Case
Now we will calculate the combinations for each case:
- For \( x = 0 \): Choose 10 from 21: \( \binom{21}{10} \)
- For \( x = 1 \): Choose 9 from 21: \( \binom{21}{9} \)
- For \( x = 2 \): Choose 8 from 21: \( \binom{21}{8} \)
- For \( x = 3 \): Choose 7 from 21: \( \binom{21}{7} \)
- For \( x = 4 \): Choose 6 from 21: \( \binom{21}{6} \)
- For \( x = 5 \): Choose 5 from 21: \( \binom{21}{5} \)
- For \( x = 6 \): Choose 4 from 21: \( \binom{21}{4} \)
- For \( x = 7 \): Choose 3 from 21: \( \binom{21}{3} \)
- For \( x = 8 \): Choose 2 from 21: \( \binom{21}{2} \)
- For \( x = 9 \): Choose 1 from 21: \( \binom{21}{1} \)
- For \( x = 10 \): Choose 0 from 21: \( \binom{21}{0} \)

### Step 6: Compute the Total
Now we can compute the total:
\[ \text{Total Ways} = \binom{21}{10} + \binom{21}{9} + \binom{21}{8} + \binom{21}{7} + \binom{21}{6} + \binom{21}{5} + \binom{21}{4} + \binom{21}{3} + \binom{21}{2} + \binom{21}{1} + \binom{21}{0} \]

### Step 7: Use the Binomial Theorem
By the binomial theorem, we know that:
\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]
Thus, for \( n = 21 \):
\[ \sum_{k=0}^{21} \binom{21}{k} = 2^{21} \]
However, since we only need half of these combinations (from \( k = 0 \) to \( k = 10 \)), we can use symmetry:
\[ \sum_{k=0}^{10} \binom{21}{k} = \frac{1}{2} \cdot 2^{21} = 2^{20} \]

### Final Answer
Thus, the total number of ways to choose 10 objects from the given set is:
\[ \text{Total Ways} = 2^{20} \]