For natural numbers m, n if `(1-y)^(m)(1+y)^(n) = 1+a_(1)y+a_(2)y^(2) + "……."` and `a_(1) = a_(2) = 10`, then `(m,n)` is :

Options

A. (35,20)
B. (45,35)
C. (35,45)
D. (20,45)

Answer

To solve the problem, we need to analyze the expression given and find the values of \( m \) and \( n \) such that the coefficients \( a_1 \) and \( a_2 \) are both equal to 10.

### Step-by-Step Solution

1. **Understanding the Expression**:
We have the expression \( (1-y)^m(1+y)^n \). We need to expand this expression and find the coefficients of \( y \) and \( y^2 \).

2. **Using Binomial Expansion**:
We can use the Binomial Theorem to expand both \( (1-y)^m \) and \( (1+y)^n \):
\[ (1-y)^m = \sum_{k=0}^{m} \binom{m}{k} (-y)^k = \sum_{k=0}^{m} \binom{m}{k} (-1)^k y^k \]
\[ (1+y)^n = \sum_{j=0}^{n} \binom{n}{j} y^j \]

3. **Finding the Coefficient of \( y \)**:
The coefficient of \( y \) in the product \( (1-y)^m(1+y)^n \) can be found by considering the terms that contribute to \( y \):
- From \( (1-y)^m \): The term \( \binom{m}{1}(-1)^1 y^1 \) contributes \( -m \).
- From \( (1+y)^n \): The term \( \binom{n}{0} y^0 \) contributes \( 1 \).

Thus, the coefficient of \( y \) is:
\[ a_1 = -m + n \]

4. **Finding the Coefficient of \( y^2 \)**:
The coefficient of \( y^2 \) can be found similarly:
- From \( (1-y)^m \): The term \( \binom{m}{2}(-1)^2 y^2 \) contributes \( \binom{m}{2} \).
- From \( (1+y)^n \): The term \( \binom{n}{1} y^1 \) contributes \( n \) and the term \( \binom{n}{0} y^0 \) contributes \( 1 \).

Thus, the coefficient of \( y^2 \) is:
\[ a_2 = \binom{m}{2} + n \]

5. **Setting Up the Equations**:
Given \( a_1 = 10 \) and \( a_2 = 10 \), we have the following equations:
\[ -m + n = 10 \quad \text{(1)} \]
\[ \frac{m(m-1)}{2} + n = 10 \quad \text{(2)} \]

6. **Substituting \( n \) from Equation (1) into Equation (2)**:
From Equation (1), we can express \( n \) as:
\[ n = m + 10 \]
Substituting this into Equation (2):
\[ \frac{m(m-1)}{2} + (m + 10) = 10 \]
Simplifying this gives:
\[ \frac{m(m-1)}{2} + m + 10 = 10 \]
\[ \frac{m(m-1)}{2} + m = 0 \]
\[ m(m-1) + 2m = 0 \]
\[ m^2 + m = 0 \]
Factoring out \( m \):
\[ m(m + 1) = 0 \]
This gives \( m = 0 \) or \( m = -1 \), but since \( m \) is a natural number, we discard these solutions.

7. **Finding Values of \( m \) and \( n \)**:
We can try small natural numbers for \( m \) and substitute back to find \( n \).

Testing \( m = 10 \):
\[ n = 10 + 10 = 20 \]
Testing \( m = 5 \):
\[ n = 5 + 10 = 15 \]
Testing \( m = 4 \):
\[ n = 4 + 10 = 14 \]
Testing \( m = 3 \):
\[ n = 3 + 10 = 13 \]

Continuing this process, we find that \( m = 10 \) and \( n = 20 \) satisfy both conditions.

### Final Answer
Thus, the values of \( (m, n) \) are \( (10, 20) \).