Question

The product of any \[r\] consecutive natural numbers is always divisible by \[r!\]
A. True
B. False

Answer 1

Hint: First of all, consider the \[r\] consecutive natural numbers as \[\left( {n + r} \right),\left( {n + r - 1} \right),...................,\left( {n + 1} \right)\]. Then find their product and simplify it further by using the formula in permutations to show that \[r!\] is a factor of \[r\] consecutive natural numbers to get the required answer.

Complete step-by-step answer:
Let us consider the \[r\] consecutive natural numbers as \[\left( {n + r} \right),\left( {n + r - 1} \right),...................,\left( {n + 1} \right)\] where \[n\] is the smallest natural number than the given \[r\] consecutive natural numbers.
Now, consider the product of these \[r\] consecutive natural numbers as
\[ \Rightarrow \left( {n + r} \right)\left( {n + r - 1} \right).........................\left( {n + 1} \right)\]
We know that \[{}^{n + r}{P_r} = \dfrac{{\left( {n + r} \right)!}}{{\left( {n + r - r} \right)!}} = \dfrac{{\left( {n + r} \right)!}}{{n!}} = \left( {n + r} \right)\left( {n + r - 1} \right).........................\left( {n + 1} \right)\].
By using this formula, the product of \[r\] consecutive natural numbers are given by
\[ \Rightarrow \left( {n + r} \right)\left( {n + r - 1} \right).........................\left( {n + 1} \right) = \dfrac{{\left( {n + r} \right)!}}{{n!}}\]
Now, if it is true that prime factors in \[\left( {n + r} \right)!\] appear just as frequently or more as in \[n!r!\], then now for some integer \[k\] that \[\left( {n + r} \right)! = k \times n! \times r!\].
So, we have \[\dfrac{{\left( {n + r} \right)!}}{{n!}} = \dfrac{{k \times n! \times r!}}{{n!}} = k \times r!\]
Hence, the product of \[r\] consecutive natural numbers are \[\left( {n + r} \right)\left( {n + r - 1} \right)............................................\left( {n + 1} \right) = k \times r!\]
Clearly, the product of \[r\] consecutive natural numbers are divisible by \[r!\] as it is a factor of the product of the \[r\] consecutive natural numbers.
Hence, proved.

So, the correct answer is “Option A”.

Note: Consecutive natural numbers are natural numbers which follow each other in the order without any gaps, from smallest to largest. For example, \[1,2,3,............\]. To solve these kinds of problems always remember the formula \[{}^{n + r}{P_r} = \dfrac{{\left( {n + r} \right)!}}{{\left( {n + r - r} \right)!}} = \dfrac{{\left( {n + r} \right)!}}{{n!}} = \left( {n + r} \right)\left( {n + r - 1} \right).........................\left( {n + 1} \right)\].