Question

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.

Answer 1

`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.