IF `""^(2n) C_3: ^n C_2=11:1` then n=

Options

A. 5
B. 4
C. 6
D. none of these

Answer

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

\[ \frac{^{2n}C_3}{^{n}C_2} = \frac{11}{1} \]

### Step 1: Write the formulas for combinations

The formula for combinations is given by:

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

Using this formula, we can express \( ^{2n}C_3 \) and \( ^{n}C_2 \):

\[ ^{2n}C_3 = \frac{(2n)!}{3!(2n-3)!} \]

\[ ^{n}C_2 = \frac{n!}{2!(n-2)!} \]

### Step 2: Set up the equation using the given ratio

Substituting these into the ratio, we have:

\[ \frac{(2n)!}{3!(2n-3)!} \div \frac{n!}{2!(n-2)!} = 11 \]

This can be rewritten as:

\[ \frac{(2n)! \cdot 2!(n-2)!}{3!(2n-3)! \cdot n!} = 11 \]

### Step 3: Cross-multiply to eliminate the fraction

Cross-multiplying gives us:

\[ (2n)! \cdot 2!(n-2)! = 11 \cdot 3!(2n-3)! \cdot n! \]

### Step 4: Substitute the values of factorials

Now we can substitute \( 2! = 2 \) and \( 3! = 6 \):

\[ (2n)! \cdot 2(n-2)! = 11 \cdot 6(2n-3)! \cdot n! \]

This simplifies to:

\[ (2n)! \cdot 2(n-2)! = 66(2n-3)! \cdot n! \]

### Step 5: Expand the factorials

We can expand \( (2n)! \) as:

\[ (2n)(2n-1)(2n-2)(2n-3)! \]

Substituting this back into the equation gives:

\[ (2n)(2n-1)(2n-2)(2n-3)! \cdot 2(n-2)! = 66(2n-3)! \cdot n! \]

### Step 6: Cancel \( (2n-3)! \)

We can cancel \( (2n-3)! \) from both sides:

\[ (2n)(2n-1)(2n-2) \cdot 2(n-2)! = 66 \cdot n! \]

### Step 7: Simplify further

Now, we can express \( n! \) as \( n(n-1)(n-2)! \):

\[ (2n)(2n-1)(2n-2) \cdot 2(n-2)! = 66 \cdot n(n-1)(n-2)! \]

Cancelling \( (n-2)! \) from both sides gives:

\[ (2n)(2n-1)(2n-2) \cdot 2 = 66n(n-1) \]

### Step 8: Rearranging the equation

Expanding both sides:

\[ 4n(2n-1)(2n-2) = 66n(n-1) \]

Dividing both sides by \( n \) (assuming \( n \neq 0 \)) gives:

\[ 4(2n-1)(2n-2) = 66(n-1) \]

### Step 9: Further simplify

Expanding both sides:

\[ 4(4n^2 - 6n + 2) = 66n - 66 \]

This simplifies to:

\[ 16n^2 - 24n + 8 = 66n - 66 \]

### Step 10: Rearranging to form a quadratic equation

Rearranging gives:

\[ 16n^2 - 90n + 74 = 0 \]

### Step 11: Solve the quadratic equation using the quadratic formula

Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 16, b = -90, c = 74 \):

\[ n = \frac{90 \pm \sqrt{(-90)^2 - 4 \cdot 16 \cdot 74}}{2 \cdot 16} \]

Calculating the discriminant:

\[ n = \frac{90 \pm \sqrt{8100 - 4736}}{32} \]
\[ n = \frac{90 \pm \sqrt{3364}}{32} \]
\[ n = \frac{90 \pm 58}{32} \]

Calculating the two possible values for \( n \):

1. \( n = \frac{148}{32} = 4.625 \) (not an integer)
2. \( n = \frac{32}{32} = 1 \)

### Conclusion

The only integer solution for \( n \) is:

\[ \boxed{1} \]