Find `1.""^20C_1 - 2.""^20C_2 + 3.""^20C_3 - …..-20.""^20C_20 = ? `

Options

A. (a) 1
B. (b) 2
C. (c) `-1`
D. (d) 0

Answer

To solve the problem \(1 \cdot \binom{20}{1} - 2 \cdot \binom{20}{2} + 3 \cdot \binom{20}{3} - \ldots - 20 \cdot \binom{20}{20}\), we can use the identity involving combinations.

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression can be rewritten using summation notation:
\[ S = \sum_{r=1}^{20} (-1)^{r-1} r \cdot \binom{20}{r} \]

2. **Using the Identity**:
We can use the identity \( r \cdot \binom{n}{r} = n \cdot \binom{n-1}{r-1} \). In our case, \( n = 20 \):
\[ r \cdot \binom{20}{r} = 20 \cdot \binom{19}{r-1} \]
Thus, we can rewrite the sum \( S \):
\[ S = \sum_{r=1}^{20} (-1)^{r-1} \cdot 20 \cdot \binom{19}{r-1} \]

3. **Factoring Out the Constant**:
Since \( 20 \) is a constant, we can factor it out of the summation:
\[ S = 20 \cdot \sum_{r=1}^{20} (-1)^{r-1} \cdot \binom{19}{r-1} \]

4. **Changing the Index of Summation**:
Let \( k = r - 1 \). When \( r = 1 \), \( k = 0 \) and when \( r = 20 \), \( k = 19 \). Therefore, we can rewrite the sum:
\[ S = 20 \cdot \sum_{k=0}^{19} (-1)^{k} \cdot \binom{19}{k} \]

5. **Using the Binomial Theorem**:
The sum \( \sum_{k=0}^{n} (-1)^k \cdot \binom{n}{k} \) is equal to \( (1 - 1)^n = 0 \) for \( n > 0 \). Here, \( n = 19 \):
\[ \sum_{k=0}^{19} (-1)^{k} \cdot \binom{19}{k} = 0 \]

6. **Final Calculation**:
Thus, substituting back into our expression for \( S \):
\[ S = 20 \cdot 0 = 0 \]

### Conclusion:
The value of the original expression is:
\[ \boxed{0} \]