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

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### 2 Answers

We have to solve for n given that C(n, 8) = C(n, 10)

C(n, 8) = C(n, 10)

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

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

=> 10*9 = (n - 8)(n - 9)

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

=> n^2 - 17n - 18= 0

=> n^2 - 18n + n - 18 = 0

=> n(n - 18) + 1(n - 18) = 0

=> (n + 1)(n - 18) = 0

=> n = -1 and n = 18

As n cannot beĀ negative take only n = 18

**The required value of n = 18**

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.**