IF `""^(28) C_(2r): "^(24) C_(2r-4)=225:11` then

Options

A. r=24
B. r=14
C. r=7
D. none of these

Answer

To solve the problem, we need to find the value of \( R \) given the ratio:

\[ \frac{28C_{2R}}{24C_{2R-4}} = \frac{225}{11} \]

### Step 1: Write the combinations in terms of factorials

Using the formula for combinations, we have:

\[ nC_r = \frac{n!}{r!(n-r)!} \]

Thus,

\[ 28C_{2R} = \frac{28!}{(2R)!(28-2R)!} \]
\[ 24C_{2R-4} = \frac{24!}{(2R-4)!(24-(2R-4))!} = \frac{24!}{(2R-4)!(28-2R)!} \]

### Step 2: Set up the equation

Substituting these into the ratio gives:

\[ \frac{\frac{28!}{(2R)!(28-2R)!}}{\frac{24!}{(2R-4)!(28-2R)!}} = \frac{225}{11} \]

This simplifies to:

\[ \frac{28!}{(2R)!(24!)} \cdot \frac{(2R-4)!}{(28-2R)!} = \frac{225}{11} \]

### Step 3: Simplify the left-hand side

The \( (28-2R)! \) cancels out, giving us:

\[ \frac{28! \cdot (2R-4)!}{(2R)! \cdot 24!} = \frac{225}{11} \]

### Step 4: Expand \( 28! \)

We can express \( 28! \) as:

\[ 28! = 28 \times 27 \times 26 \times 25 \times 24! \]

Substituting this into our equation:

\[ \frac{28 \times 27 \times 26 \times 25 \times 24! \cdot (2R-4)!}{(2R)! \cdot 24!} = \frac{225}{11} \]

The \( 24! \) cancels out:

\[ \frac{28 \times 27 \times 26 \times 25 \cdot (2R-4)!}{(2R)!} = \frac{225}{11} \]

### Step 5: Write \( (2R)! \) in terms of \( (2R-4)! \)

We can express \( (2R)! \) as:

\[ (2R)! = (2R)(2R-1)(2R-2)(2R-3)(2R-4)! \]

Substituting this into our equation gives:

\[ \frac{28 \times 27 \times 26 \times 25}{(2R)(2R-1)(2R-2)(2R-3)} = \frac{225}{11} \]

### Step 6: Cross-multiply to solve for \( R \)

Cross-multiplying gives:

\[ 28 \times 27 \times 26 \times 25 \times 11 = 225 \times (2R)(2R-1)(2R-2)(2R-3) \]

### Step 7: Calculate the left-hand side

Calculating the left-hand side:

\[ 28 \times 27 = 756 \]
\[ 756 \times 26 = 19656 \]
\[ 19656 \times 25 = 491400 \]
\[ 491400 \times 11 = 5405400 \]

So we have:

\[ 5405400 = 225 \times (2R)(2R-1)(2R-2)(2R-3) \]

### Step 8: Divide both sides by 225

Calculating:

\[ \frac{5405400}{225} = 24024 \]

So we have:

\[ (2R)(2R-1)(2R-2)(2R-3) = 24024 \]

### Step 9: Factor and solve for \( R \)

We can try \( 2R = 14 \):

\[ 14 \times 13 \times 12 \times 11 = 24024 \]

Thus, \( 2R = 14 \) implies \( R = 7 \).

### Final Answer

The value of \( R \) is:

\[ \boxed{7} \]