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 find the value of \( r \) given the equations:

\[ \frac{{^{2n}C_{2}}}{{^{n}C_{2}}} = \frac{9}{2} \]

and

\[ ^{n}C_{r} = 10. \]

### Step 1: Express the combinations
We start by expressing the combinations in terms of factorials. The combination formula is given by:

\[ ^{n}C_{r} = \frac{n!}{r!(n-r)!}. \]

Thus, we can write:

\[ ^{2n}C_{2} = \frac{(2n)!}{2!(2n-2)!} \]

and

\[ ^{n}C_{2} = \frac{n!}{2!(n-2)!}. \]

### Step 2: Set up the equation
Substituting these into the ratio gives us:

\[ \frac{\frac{(2n)!}{2!(2n-2)!}}{\frac{n!}{2!(n-2)!}} = \frac{9}{2}. \]

### Step 3: Simplify the equation
This simplifies to:

\[ \frac{(2n)!}{(2n-2)!} \cdot \frac{(n-2)!}{n!} = \frac{9}{2}. \]

### Step 4: Further simplify
Now, we can express \( (2n)! \) as:

\[ (2n)! = (2n)(2n-1)(2n-2)! \]

Thus, our equation becomes:

\[ \frac{(2n)(2n-1)(2n-2)!}{(2n-2)!} \cdot \frac{(n-2)!}{n!} = \frac{9}{2}. \]

This simplifies to:

\[ \frac{(2n)(2n-1)}{n(n-1)} = \frac{9}{2}. \]

### Step 5: Cross-multiply and solve for \( n \)
Cross-multiplying gives:

\[ 2(2n)(2n-1) = 9n(n-1). \]

Expanding both sides:

\[ 8n^2 - 4n = 9n^2 - 9n. \]

Rearranging gives:

\[ 9n^2 - 8n^2 - 9n + 4n = 0 \implies n^2 - 5n = 0. \]

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 6: Use \( n \) to find \( r \)
Now, we know \( n = 5 \) and we need to find \( r \) such that:

\[ ^{5}C_{r} = 10. \]

### Step 7: Solve for \( r \)
We know that:

\[ ^{5}C_{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10. \]

Thus, \( r = 2 \).

### Final Answer
The value of \( r \) is:

\[ \boxed{2}. \]