Question

State true or false.

The product of any $r$ consecutive natural numbers is always divisible by $r!$.

• A. True
• B. False

Answer 1

The correct option is A True

Let ${}^{\prime }{r}^{\prime }$ consecutive integers be $x+1,x+2,...x+r$

$\therefore (x+1)(x+2)...(x+r)=\frac{(x+r)(x+r-1)...(x+1)!}{x!}$

$=\frac{(x+r)!}{\left(x\right)!}.\frac{r!}{r!}{=}^{x+r}{C}_r.\left(r\right)!$

Thus, $(x+1)(x+2)...(x+r){=}^{x+r}{C}_r.\left(r\right)!$ which is clearly divisible by $\left(r\right)!$.

Hence, it is a true statement.