The number of ordered pairs (r, k) for which `6(""^(35)C_(r))=(k^(2)-3)(""^(36)C_(r+1))`, where k is an integer, is:

Options

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

Answer

To solve the equation \( 6 \binom{35}{r} = (k^2 - 3) \binom{36}{r+1} \), we will follow these steps:

### Step 1: Rewrite the Binomial Coefficient
We know that:
\[ \binom{36}{r+1} = \frac{36}{r+1} \binom{35}{r} \]
Substituting this into the equation gives:
\[ 6 \binom{35}{r} = (k^2 - 3) \frac{36}{r+1} \binom{35}{r} \]

### Step 2: Simplify the Equation
Assuming \( \binom{35}{r} \neq 0 \) (which is true for \( 0 \leq r \leq 35 \)), we can divide both sides by \( \binom{35}{r} \):
\[ 6 = (k^2 - 3) \frac{36}{r+1} \]

### Step 3: Rearranging the Equation
Rearranging gives:
\[ k^2 - 3 = \frac{6(r+1)}{36} \]
Simplifying further:
\[ k^2 - 3 = \frac{r+1}{6} \]
Thus,
\[ k^2 = \frac{r+1}{6} + 3 \]
\[ k^2 = \frac{r+1 + 18}{6} \]
\[ k^2 = \frac{r + 19}{6} \]

### Step 4: Finding Integer Values for k
For \( k^2 \) to be an integer, \( r + 19 \) must be divisible by 6. Let:
\[ r + 19 = 6m \quad \text{for some integer } m \]
This leads to:
\[ r = 6m - 19 \]

### Step 5: Determine the Range for r
Since \( r \) must be between 0 and 35 (inclusive), we set up the inequalities:
\[ 0 \leq 6m - 19 \leq 35 \]

1. **Lower Bound**:
\[ 6m - 19 \geq 0 \implies 6m \geq 19 \implies m \geq \frac{19}{6} \implies m \geq 4 \]
(since \( m \) must be an integer)

2. **Upper Bound**:
\[ 6m - 19 \leq 35 \implies 6m \leq 54 \implies m \leq 9 \]

Thus, \( m \) can take values from 4 to 9.

### Step 6: Calculate Possible Values of r
Calculating \( r \) for \( m = 4, 5, 6, 7, 8, 9 \):
- For \( m = 4 \): \( r = 6(4) - 19 = 24 - 19 = 5 \)
- For \( m = 5 \): \( r = 6(5) - 19 = 30 - 19 = 11 \)
- For \( m = 6 \): \( r = 6(6) - 19 = 36 - 19 = 17 \)
- For \( m = 7 \): \( r = 6(7) - 19 = 42 - 19 = 23 \)
- For \( m = 8 \): \( r = 6(8) - 19 = 48 - 19 = 29 \)
- For \( m = 9 \): \( r = 6(9) - 19 = 54 - 19 = 35 \)

### Step 7: Determine k for Each r
For each \( r \):
- \( k^2 = \frac{r + 19}{6} \)
- Calculate \( k \) values:
- For \( r = 5 \): \( k^2 = \frac{24}{6} = 4 \) → \( k = \pm 2 \)
- For \( r = 11 \): \( k^2 = \frac{30}{6} = 5 \) → No integer \( k \)
- For \( r = 17 \): \( k^2 = \frac{36}{6} = 6 \) → No integer \( k \)
- For \( r = 23 \): \( k^2 = \frac{42}{6} = 7 \) → No integer \( k \)
- For \( r = 29 \): \( k^2 = \frac{48}{6} = 8 \) → No integer \( k \)
- For \( r = 35 \): \( k^2 = \frac{54}{6} = 9 \) → \( k = \pm 3 \)

### Step 8: Count Ordered Pairs
The valid pairs are:
- \( (5, 2) \)
- \( (5, -2) \)
- \( (35, 3) \)
- \( (35, -3) \)

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

### Final Answer
The number of ordered pairs \( (r, k) \) is **4**.