What is the value of n if n is a positive integer?

Question

Please solve this question I am not sure how to solve this questioned provide me with the solution

If $$ 2 \times {}^n P_2 + 50 = {}^{2n}P_2 $$ where $n$ is a positive integer, then the value of $n$ is ... Options: 7,6,5,4

Answer 1

It's $n=5$.

We want to solve $2P(n,2)+50=P(2n,2)$, where $P(n,k) = \frac{n!}{(n-k)!} = \binom{n}{k}k!$. This is equivalent to solving $$2\frac{n!}{2!(n-2)!} + 25 = \frac{(2n)!}{(2n-2)!2!} \tag{A}$$ I don't see a procedural way to solve $(A)$(see edit below) to solve this (except possibly using gamma). We first note that assuming $n \ge 2$ allows every expression in the equation, in particular $(n-2)!$ and $(2n-2)!$, to be defined. I would plug in $n=2,3,4,5,6,7,8,9...$ etc until I got a solution (though I wouldn't know if it were unique) or just check wolframalpha. In your case, we luckily have this question to be multiple choice. Just plug in $n=4,5,6,7$ to see which fit. As it turns out, $n=5$ satisfies the equation.

Edit: Oh ok; I can't believe I didn't see this one. Based on Claude Leibovici's answer, the procedural way to do this is to note that

$$\frac{n!}{(n-2)!} = (n)(n-1), \frac{(2n)!}{(2n-2)!} = (2n)(2n-1) \tag{B}$$

Then use $B$ to make $A$ become $C$ as follows:

$$(n)(n-1) + 25 = (n)(2n-1) \tag{C}$$

Finally, use quadratic formula.

Answer 2

Using $P(n,k) = \frac{n!}{(n-k)!}$, you need to solve $$\frac{2 n!}{(n-2)!}-\frac{(2 n)!}{(2 n-2)!}+50=0$$ Just simplify the factorials to get $$50-n^2=0$$