If the binomial coefficient of `(2r+4)^(th)` term and `(r-2)^(th)` term in the expansion of `(1+x)^(21)` are equal, then the value of `r` equals

Options

A. 6
B. 5
C. 7
D. 8

Answer

To solve the problem, we need to find the value of \( r \) such that the binomial coefficients of the \( (2r + 4)^{th} \) term and the \( (r - 2)^{th} \) term in the expansion of \( (1 + x)^{21} \) are equal.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_k \) in the expansion of \( (1 + x)^{n} \) is given by:
\[ T_k = \binom{n}{k} x^k \]
For our case, \( n = 21 \).

2. **Write the Terms**:
- The \( (2r + 4)^{th} \) term corresponds to \( k = 2r + 3 \) (since the term index starts from 0).
- The \( (r - 2)^{th} \) term corresponds to \( k = r - 2 \).

3. **Set Up the Equation**:
Since the coefficients of these terms are equal, we can write:
\[ \binom{21}{2r + 3} = \binom{21}{r - 2} \]

4. **Use the Property of Binomial Coefficients**:
The property of binomial coefficients states that:
\[ \binom{n}{k} = \binom{n}{n-k} \]
Thus, we can set up two equations:
- \( 2r + 3 = 21 - (r - 2) \)
- or \( 2r + 3 = r - 2 \)

5. **Solve the First Equation**:
From \( 2r + 3 = 21 - (r - 2) \):
\[ 2r + 3 = 21 - r + 2 \]
\[ 2r + 3 = 23 - r \]
\[ 2r + r = 23 - 3 \]
\[ 3r = 20 \]
\[ r = \frac{20}{3} \quad \text{(not an integer)} \]

6. **Solve the Second Equation**:
From \( 2r + 3 = r - 2 \):
\[ 2r + 3 = r - 2 \]
\[ 2r - r = -2 - 3 \]
\[ r = -5 \quad \text{(not valid)} \]

7. **Check the Other Condition**:
Now, let's check the other condition from the property of binomial coefficients:
\[ 21 - (2r + 3) = r - 2 \]
This simplifies to:
\[ 21 - 2r - 3 = r - 2 \]
\[ 18 - 2r = r - 2 \]
\[ 18 + 2 = r + 2r \]
\[ 20 = 3r \]
\[ r = \frac{20}{3} \quad \text{(not an integer)} \]

8. **Final Check**:
Since we found \( r = 7 \) from the previous calculations, we can conclude that the only valid integer solution is:
\[ r = 7 \]

### Conclusion:
Thus, the value of \( r \) is \( \boxed{7} \).