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Let $5\le k<n $.Then $2k$ divides $n(n-1)(n-2)\ldots (n-k+1)$.

Case 1: $\gcd(2,n)=2$ .

For $\forall n$ either $n$ or $n-1$ is even.

Hence $2$ divides $n $ or $n-1$ and thus $2$ divides $n(n-1)(n-2)\ldots (n-k+1)$ and hence $2k$ divides $n(n-1)(n-2)\ldots (n-k+1)$.

Case 2:$\gcd(2,n)=1$ . I am unable to proceed in this case.Please give some hints.

Learnmore
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2 Answers2

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You are mistaken. You need to assume $\gcd(2,k)=1$ in Case $1$.

However, there is no need to divide cases considering that $$\binom{n}{k}=\frac{n(n-1)(n-2)\ldots (n-k+1)}{k!} \in \mathbb{Z}$$

It follows that $n(n-1)(n-2)\ldots (n-k+1)$ is divisible by $k!$.

However, since $k!=\color{blue}{k} \times (k-1) \times \dots \times \color{green}{2} \times 1$, $k!$ is divisible by $2k$.

Thus $n(n-1)(n-2)\ldots (n-k+1)$ is divisible by $2k$.

S.C.B.
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Another way is to note that product $n(n-1)(n-2) \cdots (n-k+1)$ is a product of $k$ consecutive positive numbers and hence $k$ divides at least one factor. If $k\geq 4$ then among the other factors there is at least one even number. Hence $2k$ divides $n(n-1)(n-2) \cdots (n-k+1)$.

P.S. The statement is false for $k=2$, (take for example $n=3$). For $k=3$, it holds because $2k=6$ divides $n(n-1)(n-2)$ (try to prove it).

Robert Z
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  • Why is a product of $k$ consecutive numbers divisible by $k$ – Learnmore Aug 13 '16 at 07:55
  • @S.Bandopadhaya Assume that the smallest number is $qk+r$ with $r$ the remainder of the division by $k$, then the $k$ factors have the form $qk+(r+j)$ for $j=0,\dots,k-1$. Hence the factor $qk+(r+j)$ with $j=k-r\in [0,k-1]$ will be divisible by $k$. Also have a look here: http://math.stackexchange.com/questions/126766/prove-that-one-of-n-consecutive-integers-must-be-divisible-by-n – Robert Z Aug 13 '16 at 08:02