To solve the problem where \( C(n, 10) = C(n, 12) \) and we need to determine \( C(n, 5) \), we will follow these steps:
### Step 1: Set up the equation
We know that \( C(n, k) = \frac{n!}{k!(n-k)!} \). Therefore, we can write:
\[
C(n, 10) = \frac{n!}{10!(n-10)!}
\]
\[
C(n, 12) = \frac{n!}{12!(n-12)!}
\]
Given that \( C(n, 10) = C(n, 12) \), we can set these two expressions equal to each other:
\[
\frac{n!}{10!(n-10)!} = \frac{n!}{12!(n-12)!}
\]
### Step 2: Cancel \( n! \)
Since \( n! \) appears on both sides, we can cancel it (assuming \( n! \neq 0 \)):
\[
\frac{1}{10!(n-10)!} = \frac{1}{12!(n-12)!}
\]
### Step 3: Cross-multiply
Cross-multiplying gives us:
\[
12!(n-12)! = 10!(n-10)!
\]
### Step 4: Rewrite factorials
We can rewrite \( 12! \) and \( 10! \) in terms of each other:
\[
12! = 12 \times 11 \times 10!
\]
So, substituting this into the equation gives:
\[
12 \times 11 \times 10!(n-12)! = 10!(n-10)!
\]
### Step 5: Cancel \( 10! \)
Now we can cancel \( 10! \) from both sides:
\[
12 \times 11 \times (n-12)! = (n-10)!
\]
### Step 6: Rewrite \( (n-10)! \)
We can express \( (n-10)! \) in terms of \( (n-12)! \):
\[
(n-10)! = (n-10)(n-11)(n-12)!
\]
Substituting this into the equation gives:
\[
12 \times 11 \times (n-12)! = (n-10)(n-11)(n-12)!
\]
### Step 7: Cancel \( (n-12)! \)
Assuming \( (n-12)! \neq 0 \), we can cancel it:
\[
12 \times 11 = (n-10)(n-11)
\]
### Step 8: Expand and rearrange
Expanding the right side:
\[
12 \times 11 = n^2 - 21n + 110
\]
Setting the equation:
\[
n^2 - 21n + 110 - 132 = 0
\]
This simplifies to:
\[
n^2 - 21n - 22 = 0
\]
### Step 9: Solve the quadratic equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
n = \frac{21 \pm \sqrt{(-21)^2 - 4 \cdot 1 \cdot (-22)}}{2 \cdot 1}
\]
\[
n = \frac{21 \pm \sqrt{441 + 88}}{2}
\]
\[
n = \frac{21 \pm \sqrt{529}}{2}
\]
\[
n = \frac{21 \pm 23}{2}
\]
This gives us two possible values for \( n \):
\[
n = 22 \quad \text{or} \quad n = -1
\]
Since \( n \) must be a positive integer, we take \( n = 22 \).
### Step 10: Calculate \( C(n, 5) \)
Now we need to find \( C(22, 5) \):
\[
C(22, 5) = \frac{22!}{5!(22-5)!} = \frac{22!}{5! \cdot 17!}
\]
Calculating this gives:
\[
C(22, 5) = \frac{22 \times 21 \times 20 \times 19 \times 18}{5 \times 4 \times 3 \times 2 \times 1} = 26334
\]
### Final Answer
Thus, the value of \( C(n, 5) \) is:
\[
\boxed{26334}
\]
