IF `""^n C_6: "^(n-3)C_3=33:4` then n=

Options

A. 9
B. 10
C. 11
D. none of these

Answer

To solve the equation \( \binom{n}{6} : \binom{n-3}{3} = 33 : 4 \), we will follow these steps:

### Step 1: Write the combinations in terms of factorials
The combination formula is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
So we can write:
\[ \binom{n}{6} = \frac{n!}{6!(n-6)!} \]
and
\[ \binom{n-3}{3} = \frac{(n-3)!}{3!(n-6)!} \]

### Step 2: Set up the ratio
Now we can substitute these into the ratio:
\[ \frac{\binom{n}{6}}{\binom{n-3}{3}} = \frac{\frac{n!}{6!(n-6)!}}{\frac{(n-3)!}{3!(n-6)!}} = \frac{n! \cdot 3!}{6! \cdot (n-3)!} \]

### Step 3: Simplify the expression
The \( (n-6)! \) cancels out:
\[ = \frac{n! \cdot 6}{6! \cdot (n-3)!} = \frac{n(n-1)(n-2)(n-3)! \cdot 6}{6! \cdot (n-3)!} \]
Now, we can cancel \( (n-3)! \):
\[ = \frac{n(n-1)(n-2) \cdot 6}{720} \]
Since \( 6! = 720 \).

### Step 4: Set the ratio equal to \( \frac{33}{4} \)
Now we have:
\[ \frac{n(n-1)(n-2) \cdot 6}{720} = \frac{33}{4} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 4n(n-1)(n-2) \cdot 6 = 33 \cdot 720 \]
Calculating \( 33 \cdot 720 \):
\[ 33 \cdot 720 = 23760 \]
So we have:
\[ 24n(n-1)(n-2) = 23760 \]

### Step 6: Divide both sides by 24
\[ n(n-1)(n-2) = \frac{23760}{24} = 990 \]

### Step 7: Solve for \( n \)
Now we need to find \( n \) such that:
\[ n(n-1)(n-2) = 990 \]
Let's test some integer values for \( n \):
- For \( n = 10 \):
\[ 10 \cdot 9 \cdot 8 = 720 \quad (\text{too low}) \]
- For \( n = 11 \):
\[ 11 \cdot 10 \cdot 9 = 990 \quad (\text{correct}) \]

Thus, \( n = 11 \).

### Final Answer
\[ \boxed{11} \]