If `.^(2n)C_(3):^(n)C_(2)=44:3`, for which of the following value of `r`, the value of `.^(n)C_(r)` will be 15.

Options

A. r=3
B. r=4
C. r=5
D. r=6

Answer

To solve the problem, we need to analyze the given equation and find the appropriate value of \( r \) such that \( \binom{n}{r} = 15 \).

### Step 1: Set up the equation
We start with the equation given in the problem:
\[ \frac{\binom{2n}{3}}{\binom{n}{2}} = \frac{44}{3} \]

### Step 2: Express the combinations in factorial form
Using the definition of combinations, we can express both sides in terms of factorials:
\[ \frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{2!(n-2)!}} = \frac{44}{3} \]

### Step 3: Simplify the equation
This simplifies to:
\[ \frac{(2n)! \cdot 2! \cdot (n-2)!}{3! \cdot (2n-3)! \cdot n!} = \frac{44}{3} \]
Substituting \( 2! = 2 \) and \( 3! = 6 \):
\[ \frac{(2n)! \cdot 2 \cdot (n-2)!}{6 \cdot (2n-3)! \cdot n!} = \frac{44}{3} \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 3 \cdot (2n)! \cdot 2 \cdot (n-2)! = 44 \cdot 6 \cdot (2n-3)! \cdot n! \]
This simplifies to:
\[ 6 \cdot (2n)! \cdot (n-2)! = 264 \cdot (2n-3)! \cdot n! \]

### Step 5: Expand factorials
Using the factorial definitions:
\[ (2n)! = (2n)(2n-1)(2n-2)(2n-3)! \]
Substituting this into the equation:
\[ 6 \cdot (2n)(2n-1)(2n-2)(2n-3)! \cdot (n-2)! = 264 \cdot (2n-3)! \cdot n! \]

### Step 6: Cancel \( (2n-3)! \)
Cancelling \( (2n-3)! \) from both sides gives:
\[ 6 \cdot (2n)(2n-1)(2n-2)(n-2)! = 264 \cdot n! \]

### Step 7: Simplify further
We can express \( n! \) as \( n \cdot (n-1)(n-2)! \):
\[ 6 \cdot (2n)(2n-1)(2n-2) = 264 \cdot n \cdot (n-1) \]

### Step 8: Divide both sides by 6
Dividing both sides by 6:
\[ (2n)(2n-1)(2n-2) = 44 \cdot n \cdot (n-1) \]

### Step 9: Solve for \( n \)
Expanding both sides and simplifying will lead us to find \( n \). After solving, we find:
\[ n = 6 \]

### Step 10: Find \( r \) such that \( \binom{n}{r} = 15 \)
Now we need to find \( r \) such that:
\[ \binom{6}{r} = 15 \]

### Step 11: Check possible values of \( r \)
Calculating \( \binom{6}{r} \) for possible values of \( r \):
- \( r = 0: \binom{6}{0} = 1 \)
- \( r = 1: \binom{6}{1} = 6 \)
- \( r = 2: \binom{6}{2} = 15 \) (This is a match)
- \( r = 3: \binom{6}{3} = 20 \)
- \( r = 4: \binom{6}{4} = 15 \) (This is also a match)
- \( r = 5: \binom{6}{5} = 6 \)
- \( r = 6: \binom{6}{6} = 1 \)

### Conclusion
The values of \( r \) for which \( \binom{6}{r} = 15 \) are \( r = 2 \) and \( r = 4 \).