Use combinatorics to solve the equation (n,8)=(n,10).

We'll recall the formula that gives the number of
combinations of n elements taken k at a time:

(n,k) =
n!/k!*(n-k)!

Now, we'll apply this formula to write (n,8)
and (n,10) in factorial way:

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

(n,10) = n!/10!*(n-10)!
(2)

We'll equate (1) and
(2):

n!/8!*(n-8)! =
n!/10!*(n-10)!

But (n-8)! =
(n-10)!*(n-9)*(n-8)

10! =
8!*9*10

n!/8!*(n-10)!*(n-9)*(n-8) =
n!/8!*9*10*(n-10)!

We'll simplify and we'll
get:

1/(n-9)*(n-8) =
1/90

We'll cross
multiply:

(n-9)*(n-8) =
90

We'll remove the
brackets:

n^2 - 17n + 72 - 90 =
0

n^2 - 17n - 18 = 0

We'll
apply quadratic formula;

n1 = [17 + sqrt(289 +
72)]/2

n1 = (17+19)/2

n1 =
18

n2 = -1

Since
n has to be a natural number, then we'll keep the positive natural value: n =
18.