If the coefficient of (3r + 4)th term and (2r - 2)th term in the expansion of `(1 + x)^(20)` are equal then r is equal to,

Options

A. 4
B. 5
C. 7
D. 8

Answer

To solve the problem, we need to find the value of \( r \) such that the coefficients of the \( (3r + 4) \)th term and the \( (2r - 2) \)th term in the expansion of \( (1 + x)^{20} \) are equal.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_k \) in the expansion of \( (1 + x)^{20} \) is given by:
\[ T_k = \binom{20}{k} x^k \]
where \( k \) is the term number starting from 0.

2. **Find the Coefficient of the \( (3r + 4) \)th Term**:
The \( (3r + 4) \)th term corresponds to \( k = 3r + 4 - 1 = 3r + 3 \). Thus, the coefficient of this term is:
\[ C_1 = \binom{20}{3r + 3} \]

3. **Find the Coefficient of the \( (2r - 2) \)th Term**:
The \( (2r - 2) \)th term corresponds to \( k = 2r - 2 - 1 = 2r - 3 \). Thus, the coefficient of this term is:
\[ C_2 = \binom{20}{2r - 3} \]

4. **Set the Coefficients Equal**:
According to the problem, these coefficients are equal:
\[ \binom{20}{3r + 3} = \binom{20}{2r - 3} \]

5. **Use the Properties of Binomial Coefficients**:
The equality of binomial coefficients gives us two cases:
- Case 1: \( 3r + 3 = 2r - 3 \)
- Case 2: \( 3r + 3 + (2r - 3) = 20 \)

6. **Solve Case 1**:
\[ 3r + 3 = 2r - 3 \]
Rearranging gives:
\[ 3r - 2r = -3 - 3 \implies r = -6 \]
Since \( r \) cannot be negative, we discard this case.

7. **Solve Case 2**:
\[ 3r + 3 + 2r - 3 = 20 \]
Simplifying gives:
\[ 5r = 20 \implies r = 4 \]

8. **Conclusion**:
The only valid solution is:
\[ r = 4 \]

### Final Answer:
Thus, the value of \( r \) is \( 4 \).