Answer to Question #120840 in Combinatorics | Number Theory for Hussain Ousman Darboe

Question

nP8+nP13+nP20+1=nP9+nP15+4 where P means permutations, solve for n?

Answer 1

There is something wrong with this question. I tried to solve it but found that the equation does not hold for a value of n, given that n is an integer and $n \geq 20$ .

Basically, I checked if there is any value of n which satisfies this equation. For this I calculated the value of the function f(n) = nP8 + nP13 + nP20 + 1- nP9 - nP15 - 4 for n = 20, 21, 22, ....

I found that f(n) = 2.41311 x 10¹⁸ for n = 20 and f(n) is a monotonically increasing function of n. Hence, I did not find any value of n for which f(n) = 0.

Hence, the equation does not hold for any value of n.

The problem seemed to be a simple one, that's why I sent it for approval. Obviously, I did not solve the question before getting the start working notification. When after getting the notification I started working, it seemed to be incorrect. Then I chatted with the operator who asked me to provide with the details in here.Here is the graph of the function. Since the value of f(n) are varying abruptly high, so we have used logarithmic scale in the vertical axis. Clearly, it is showing a monotonically increasing trend.

Some of the initial values of (n, f(n)) are also provided here:

The given equation is:

nP8+nP13+nP20+1=nP9+nP15+4

If each term of the equation is rewritten using factorial function, then the equation becomes

$\frac{n!}{(n-8)!}+\frac{n!}{(n-13)!}+\frac{n!}{(n-20)!}+1=\frac{n!}{(n-9)!}+\frac{n!}{(n-15)!}+4$