If C(n,9)=C(n,8) then C(n,17) is

To solve this problem, we need to use the concept of combinations, denoted by C(n, r), which represents the number of ways to choose r items from a set of n items without considering their order.

Given that C(n, 9) is equal to C(n, 8), we can set up the following equation:

C(n, 9) = C(n, 8)

To simplify this equation, we can use the formula for combinations:

C(n, r) = n! / (r! * (n-r)!)

Applying this formula to both sides of the equation, we get:

n! / (9! * (n-9)!) = n! / (8! * (n-8)!)

Next, we can simplify the equation by canceling out the common terms:

1 / (9! * (n-9)!) = 1 / (8! * (n-8)!)

Since the denominators are equal, we can conclude that the numerators must also be equal:

9! * (n-9)! = 8! * (n-8)!

Now, we can simplify the equation further by canceling out the common terms:

(n-9)! = (n-8)!

Since the factorials are equal, we can conclude that (n-9) must be equal to (n-8):

n - 9 = n - 8

Simplifying this equation, we get:

-9 = -8

This equation is not true, which means that there is no value of n that satisfies the given condition.

Therefore, we can conclude that if C(n, 9) is equal to C(n, 8), then C(n, 17) is not defined.