Question

The product of $n$ consecutive natural numbers is always divisible by

• A. $4n!$
• B. $3n!$
• C. $2n!$
• D. $n!$

Answer 1

The correct option is D $n!$

Let $n$ consecutive natural numbers be $(k+1),(k+2)........(k+n)$, where $k\in I>0$
Then product of these numbers are $=(k+1)(k+2)........(k+n)=P$ (say)
$\Rightarrow P=\frac{k!(k+1)(k+2)........(k+n)}{k!}=\frac{(k+n)!}{k!}=\frac{(k+n)!}{k!n!}(n!)=(n!){\cdot }^{n+k}{C}_n$
Hence, product of $n$ natural numbers will be always divisible by $n!$.