If `.^36C_(r+1) xx (k^2-3) = 6 xx .^35C_r ` , then number of ordered pairs (r, k) are (where `kinI`).

Options

A. 1. 4
B. 2. 6
C. 3. 2
D. 4. 3

Answer

To solve the equation given in the problem, we start with the expression:

\[ ^{36}C_{r+1} \cdot (k^2 - 3) = 6 \cdot ^{35}C_r \]

### Step 1: Apply the combination identity
Using the identity \( ^nC_r = \frac{n}{r} \cdot ^{n-1}C_{r-1} \), we can express \( ^{36}C_{r+1} \) in terms of \( ^{35}C_r \):

\[ ^{36}C_{r+1} = \frac{36}{r+1} \cdot ^{35}C_r \]

### Step 2: Substitute into the equation
Substituting this back into the original equation gives:

\[ \frac{36}{r+1} \cdot ^{35}C_r \cdot (k^2 - 3) = 6 \cdot ^{35}C_r \]

### Step 3: Cancel \( ^{35}C_r \)
Assuming \( ^{35}C_r \neq 0 \), we can divide both sides by \( ^{35}C_r \):

\[ \frac{36}{r+1} \cdot (k^2 - 3) = 6 \]

### Step 4: Simplify the equation
Multiplying both sides by \( r + 1 \) gives:

\[ 36(k^2 - 3) = 6(r + 1) \]

Dividing both sides by 6:

\[ 6(k^2 - 3) = r + 1 \]

### Step 5: Rearranging the equation
Rearranging this gives:

\[ k^2 - 3 = \frac{r + 1}{6} \]

Thus, we can express \( k^2 \) as:

\[ k^2 = \frac{r + 1}{6} + 3 \]

### Step 6: Finding integer values for \( k \)
For \( k \) to be an integer, \( \frac{r + 1}{6} + 3 \) must be a perfect square. This implies that \( r + 1 \) must be divisible by 6. Let \( r + 1 = 6m \) for some integer \( m \), then:

\[ k^2 = m + 3 \]

### Step 7: Analyzing the equation
Now we need \( m + 3 \) to be a perfect square. Let \( k^2 = n^2 \) for some integer \( n \):

\[ m = n^2 - 3 \]

### Step 8: Finding valid \( m \)
Since \( m \) must be non-negative, we have:

\[ n^2 - 3 \geq 0 \implies n^2 \geq 3 \implies n \geq 2 \]

### Step 9: Finding values for \( r \)
Substituting \( m \) back, we have:

\[ r + 1 = 6(n^2 - 3) \implies r = 6(n^2 - 3) - 1 \]

### Step 10: Finding integer pairs
Now we can find pairs of \( (r, k) \):
1. For \( n = 2 \):
- \( m = 2^2 - 3 = 1 \)
- \( r = 6(1) - 1 = 5 \)
- \( k = \pm 2 \) (since \( k^2 = 4 \))

2. For \( n = 3 \):
- \( m = 3^2 - 3 = 6 \)
- \( r = 6(6) - 1 = 35 \)
- \( k = \pm 3 \) (since \( k^2 = 9 \))

### Conclusion
The valid ordered pairs \( (r, k) \) are:
- \( (5, 2) \)
- \( (5, -2) \)
- \( (35, 3) \)
- \( (35, -3) \)

Thus, the total number of ordered pairs \( (r, k) \) is **4**.