If C ( 2n,r) = C(2n,r+2) , find r in term of n.

Answer

To solve the equation \( C(2n, r) = C(2n, r + 2) \), we will follow these steps:

### Step 1: Write the Combinations in Factorial Form
We start by expressing the combinations in terms of factorials:
\[ C(2n, r) = \frac{(2n)!}{r!(2n - r)!} \]
\[ C(2n, r + 2) = \frac{(2n)!}{(r + 2)!(2n - (r + 2))!} \]

### Step 2: Set the Two Expressions Equal
Since \( C(2n, r) = C(2n, r + 2) \), we can set the two expressions equal:
\[ \frac{(2n)!}{r!(2n - r)!} = \frac{(2n)!}{(r + 2)!(2n - r - 2)!} \]

### Step 3: Cancel Out the Factorials
We can cancel \( (2n)! \) from both sides:
\[ \frac{1}{r!(2n - r)!} = \frac{1}{(r + 2)!(2n - r - 2)!} \]

### Step 4: Cross Multiply
Cross multiplying gives us:
\[ (2n - r - 2)! \cdot 1 = (r + 2)(r + 1) \cdot (2n - r)! \]

### Step 5: Rewrite the Factorials
We can rewrite \( (r + 2)! \) as \( (r + 2)(r + 1)(r!) \) and \( (2n - r)! \) as \( (2n - r)(2n - r - 1)(2n - r - 2)! \):
\[ (2n - r - 2)! = (r + 2)(r + 1)(2n - r)(2n - r - 1)(2n - r - 2)! \]

### Step 6: Cancel the Common Factorials
Cancelling \( (2n - r - 2)! \) from both sides, we have:
\[ 1 = (r + 2)(r + 1)(2n - r)(2n - r - 1) \]

### Step 7: Expand and Rearrange
Expanding the left side:
\[ (r + 2)(r + 1) = r^2 + 3r + 2 \]
And the right side:
\[ (2n - r)(2n - r - 1) = 4n^2 - 4nr + r^2 + r \]

### Step 8: Set the Equation
Setting the two sides equal gives:
\[ r^2 + 3r + 2 = 4n^2 - 4nr + r^2 + r \]

### Step 9: Simplify
Cancelling \( r^2 \) from both sides:
\[ 3r + 2 = 4n^2 - 4nr + r \]
Rearranging gives:
\[ 2r + 4nr = 4n^2 - 2 \]

### Step 10: Factor Out r
Factoring out \( r \):
\[ r(2 + 4n) = 4n^2 - 2 \]

### Step 11: Solve for r
Finally, we solve for \( r \):
\[ r = \frac{4n^2 - 2}{2 + 4n} \]

### Step 12: Simplify Further
This can be simplified to:
\[ r = \frac{2(2n^2 - 1)}{2(1 + 2n)} = \frac{2n^2 - 1}{1 + 2n} \]

Thus, the final answer is:
\[ r = n - 1 \]