Solve (n!)/(8!(n-8)!) = (n!)/(16!(n-16)!) algebraically?

I got a solution of n=21, but I had to use the guess and check method.

1 Answer
May 30, 2018

n=24

Explanation:

We have

(n!)/(8!(n-8)!)=(n!)/(16!(n-16)!)

As n! is nonzero, we can divide both parts by it:

cancel(n!)/(8!(n-8)!)=cancel(n!)/(16!(n-16)!)

Leaving us with a much simpler equation:

1/(8!(n-8)!)=1/(16!(n-16)!)

=> 8!(n-8)! = 16!(n-16)!

We can write this as:

((n-8)!)/((n-16)!)=(16!)/(8!)

One clear solution of this is obtained by letting n be the solution to the system of equations

{(n-8=16),(n-16=8) :}

From which we get that n=24.

But is this the only solution? To check this, let "A"_(n)^k=(n!)/(k!), k<=n.

Through the definition of a factorial, "A"_n^k is equivalent to

"A"_n^k = n(n-1)(n-2)...(k+2)(k+1)

Using this new notation:

"A"_((n-8))^((n-16))="A"_16^8

(n-8)(n-9)(n-10)...(n-14)(n-15)=16*...*10*9

We can see that the left hand side product has exactly 8 terms and so does the right hand side one. All of these terms are consecutive positive integers, and as a result, they must be equal in pairs; one by one.

{(n-8=16),(n-9=15),(vdots),(n-15=9) :}

From which we get that n = 24, proving it to be the unique solution.