Statement 1: Number of ways of selecting 10 objects from 42 objects of which 21 objects are identical and remaining objects are distinct is `2^(20)dot` Statement 2: `^42 C_0+^(42)C_1+^(42)C_2++^(42)C_(21)=2^(41)dot`

Explanation

To solve the problem, we will analyze both statements step by step.

### Step 1: Analyze Statement 1
**Statement 1:** The number of ways of selecting 10 objects from 42 objects, of which 21 are identical and the remaining are distinct, is \(2^{20}\).

1. **Identify the objects:** We have 21 identical objects and 21 distinct objects, totaling 42 objects.
2. **Selection possibilities:** We can select \(x\) identical objects (where \(x\) can range from 0 to 10) and \(10 - x\) distinct objects. The number of ways to select \(x\) identical objects is always 1 (since they are identical).
3. **Calculate combinations:** The number of ways to select \(10 - x\) distinct objects from 21 distinct objects is given by \(\binom{21}{10 - x}\).

### Step 2: Set up the equation
We will sum the combinations for all possible values of \(x\):
\[ \text{Total ways} = \sum_{x=0}^{10} \binom{21}{10 - x} \]
This can be rewritten as:
\[ \text{Total ways} = \binom{21}{10} + \binom{21}{9} + \binom{21}{8} + \ldots + \binom{21}{0} \]

### Step 3: Use the Binomial Theorem
According to the Binomial Theorem:
\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]
For our case, we need the sum from \(k=0\) to \(k=10\):
\[ \sum_{k=0}^{10} \binom{21}{k} = 2^{21} \]
However, we only need half of this sum, since we are only selecting up to 10 objects from 21. Thus:
\[ \sum_{k=0}^{10} \binom{21}{k} = 2^{20} \]

### Conclusion for Statement 1
Thus, Statement 1 is correct: The number of ways to select 10 objects from 42 objects (21 identical and 21 distinct) is indeed \(2^{20}\).

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### Step 4: Analyze Statement 2
**Statement 2:** \(\binom{42}{0} + \binom{42}{1} + \binom{42}{2} + \ldots + \binom{42}{21} = 2^{41}\).

1. **Total combinations:** According to the Binomial Theorem:
\[ \sum_{k=0}^{42} \binom{42}{k} = 2^{42} \]
2. **Half of the combinations:** The sum from \(k=0\) to \(k=21\) is half of the total combinations:
\[ \sum_{k=0}^{21} \binom{42}{k} = \frac{1}{2} \cdot 2^{42} = 2^{41} \]

### Conclusion for Statement 2
Thus, Statement 2 is also correct: The sum of the combinations from 0 to 21 is indeed \(2^{41}\).

### Final Answer
- **Statement 1:** Correct
- **Statement 2:** Correct

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