if `.^(2n)C_(2):^(n)C_(2)=9:2 and .^(n)C_(r)=10`, then r is equal to

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to work with the given equations step by step.

### Step 1: Set up the equation
We are given the ratio:
\[ \frac{^{2n}C_{2}}{^{n}C_{2}} = \frac{9}{2} \]
This can be rewritten using the formula for combinations:
\[ ^{2n}C_{2} = \frac{(2n)!}{2!(2n-2)!} \quad \text{and} \quad ^{n}C_{2} = \frac{n!}{2!(n-2)!} \]
Thus, we can express the ratio as:
\[ \frac{\frac{(2n)!}{2!(2n-2)!}}{\frac{n!}{2!(n-2)!}} = \frac{9}{2} \]

### Step 2: Simplify the equation
Cancelling \(2!\) from both the numerator and denominator, we get:
\[ \frac{(2n)!}{(2n-2)!} \cdot \frac{(n-2)!}{n!} = \frac{9}{2} \]
This simplifies to:
\[ \frac{(2n)(2n-1)}{n(n-1)} = \frac{9}{2} \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ 2(2n)(2n-1) = 9n(n-1) \]
Expanding both sides:
\[ 4n(2n-1) = 9n(n-1) \]
This simplifies to:
\[ 8n^2 - 4n = 9n^2 - 9n \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 8n^2 - 9n^2 + 9n - 4n = 0 \]
Which simplifies to:
\[ -n^2 + 5n = 0 \]
Factoring out \(n\):
\[ n(n - 5) = 0 \]
Thus, \(n = 0\) or \(n = 5\). Since \(n\) must be a positive integer, we have:
\[ n = 5 \]

### Step 5: Use the value of n to find r
Now we use the second part of the problem, which states:
\[ ^{n}C_{r} = 10 \]
Substituting \(n = 5\):
\[ ^{5}C_{r} = 10 \]
Using the combination formula:
\[ \frac{5!}{r!(5-r)!} = 10 \]

### Step 6: Solve for r
We know that \(^{5}C_{2} = 10\), which implies:
\[ r = 2 \]

Thus, the value of \(r\) is:
\[ \boxed{2} \]