IF `""^15 C_r: "^15 C_(r-1)=11:5` then r=

Options

A. 4
B. 5
C. 6
D. 7

Answer

To solve the problem \( \binom{15}{r} : \binom{15}{r-1} = \frac{11}{5} \), we will use the properties of combinations.

### Step-by-step Solution:

1. **Write the ratio of combinations**:
\[ \frac{\binom{15}{r}}{\binom{15}{r-1}} = \frac{11}{5} \]

2. **Use the formula for combinations**:
The formula for combinations is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Therefore, we can express the combinations as:
\[ \binom{15}{r} = \frac{15!}{r!(15-r)!} \]
\[ \binom{15}{r-1} = \frac{15!}{(r-1)!(15-(r-1))!} = \frac{15!}{(r-1)!(16-r)!} \]

3. **Substitute the combinations into the ratio**:
\[ \frac{\frac{15!}{r!(15-r)!}}{\frac{15!}{(r-1)!(16-r)!}} = \frac{11}{5} \]

4. **Simplify the equation**:
The \( 15! \) cancels out:
\[ \frac{(16-r)!}{(15-r)!} \cdot \frac{(r-1)!}{r!} = \frac{11}{5} \]
This simplifies to:
\[ \frac{(16-r)}{r} = \frac{11}{5} \]

5. **Cross-multiply**:
\[ 5(16 - r) = 11r \]

6. **Expand and rearrange the equation**:
\[ 80 - 5r = 11r \]
\[ 80 = 11r + 5r \]
\[ 80 = 16r \]

7. **Solve for \( r \)**:
\[ r = \frac{80}{16} = 5 \]

### Final Answer:
Thus, the value of \( r \) is \( 5 \).

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