If `""^(n)C_(7)=""^(n)C_(5)`, find `""^(n)C_(4)`.

Answer

To solve the problem where \( \binom{n}{7} = \binom{n}{5} \) and find \( \binom{n}{4} \), we will follow these steps:

### Step 1: Set up the equation using the combination formula
The combination formula is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Using this formula, we can express \( \binom{n}{7} \) and \( \binom{n}{5} \):
\[ \binom{n}{7} = \frac{n!}{7!(n-7)!} \]
\[ \binom{n}{5} = \frac{n!}{5!(n-5)!} \]

### Step 2: Set the two combinations equal to each other
Since we know \( \binom{n}{7} = \binom{n}{5} \), we can set the two expressions equal:
\[ \frac{n!}{7!(n-7)!} = \frac{n!}{5!(n-5)!} \]

### Step 3: Cancel \( n! \) from both sides
Assuming \( n! \neq 0 \), we can cancel \( n! \) from both sides:
\[ \frac{1}{7!(n-7)!} = \frac{1}{5!(n-5)!} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ 5!(n-5)! = 7!(n-7)! \]

### Step 5: Substitute \( 7! \) in terms of \( 5! \)
We know that \( 7! = 7 \times 6 \times 5! \), so we can substitute this into the equation:
\[ 5!(n-5)! = 7 \times 6 \times 5!(n-7)! \]

### Step 6: Cancel \( 5! \) from both sides
Assuming \( 5! \neq 0 \), we can cancel \( 5! \):
\[ (n-5)! = 42(n-7)! \]

### Step 7: Express \( (n-5)! \) in terms of \( (n-7)! \)
We can express \( (n-5)! \) as:
\[ (n-5)(n-6)(n-7)! = 42(n-7)! \]
Now, we can cancel \( (n-7)! \) from both sides:
\[ (n-5)(n-6) = 42 \]

### Step 8: Expand and rearrange the equation
Expanding gives us:
\[ n^2 - 11n + 30 - 42 = 0 \]
which simplifies to:
\[ n^2 - 11n - 12 = 0 \]

### Step 9: Factor the quadratic equation
Now we can factor the quadratic:
\[ (n - 12)(n + 1) = 0 \]
Thus, \( n = 12 \) or \( n = -1 \). Since \( n \) must be a non-negative integer, we take \( n = 12 \).

### Step 10: Calculate \( \binom{n}{4} \)
Now we need to find \( \binom{12}{4} \):
\[ \binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12!}{4! \cdot 8!} \]
Calculating this gives:
\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \frac{11880}{24} = 495 \]

### Final Answer
Thus, \( \binom{n}{4} = 495 \).

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