If $2k$ divides $P(n,k)$? Show that for $5\le k <n$, $2k$ divides $n(n-1)(n-2)..............(n-k+1)$
Answer:- My Attempt:- The product $n(n-1)(n-2)..............(n-k+1)= ^nP_k$
We know, $^nP_k=k\,! \times ^nC_k$
Since $5\le k $, $2$ is always a divisor of $k\,! $. 
Again $k$ itself is a divisor of $k\,!$.
Thus $2k$ divides $k\,!$. And, we know $^nC_k$ is always a natural number. 
Hence the result. 
Is this the correct approach? In the answer sheet we were asked to write whether the above statement is true or not with a proper reason. I am afraid if this is the correct explanation.
 A: You have some errors.

$2$ is a divisor of $k!$, not a multiple of $k!$.

Similarly, $k$ is a divisor of $k!$, not a multiple of $k!$.

Also, it's not automatic that the product of two divisors of $k!$ is a divisor of $k!$.

Of course, in this case, assuming $k > 2$, it's clear that $2$ and $k$ are distinct factors of $(1)(2)\cdots(k)$, so you do get that $2k$ is a divisor of  $k!$.

Using the basic ideas of your argument, the proof could be written as follows . . .

Let $x=P(n,k)$.

We want to show $2k{\,\mid\,}x$.

As you correctly showed, $x$ is a multiple of $k!$.

Thus, we can write $x=(k!)y$, for some positive integer $y$.

Then 
$$\frac{x}{2k}=\frac{(k!)y}{2k}=\frac{((k-1)!)y}{2}$$
hence, since $(k-1)!$ is even, we get that ${\large{\frac{x}{2k}}}$ is a positive integer.

It follows that $2k{\,\mid\,}x$, as was to be shown.

Notes:


*

*The above argument assumes $k\ge 3$, so as to guarantee the evenness of $(k-1)!$.

*If $k=2$, and $n=3$, then $x = 6$, and $2k=4$, so in that case, $x$ is not a multiple of $2k$.

*Also, we can allow $n=k$, hence we can relax the inequality to $3 \le k \le n$.

