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

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

Explanation

Solution

First of all, consider the rr consecutive natural numbers as (n+r),(n+r−1),...................,(n+1)\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!r! is a factor of rr consecutive natural numbers to get the required answer.

Complete step-by-step answer:
Let us consider the rr consecutive natural numbers as (n+r),(n+r−1),...................,(n+1)\left( {n + r} \right),\left( {n + r - 1} \right),...................,\left( {n + 1} \right) where nn is the smallest natural number than the given rr consecutive natural numbers.
Now, consider the product of these rr consecutive natural numbers as
⇒(n+r)(n+r−1).........................(n+1)\Rightarrow \left( {n + r} \right)\left( {n + r - 1} \right).........................\left( {n + 1} \right)
We know that n+rPr=(n+r)!(n+r−r)!=(n+r)!n!=(n+r)(n+r−1).........................(n+1){}^{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 rr consecutive natural numbers are given by
⇒(n+r)(n+r−1).........................(n+1)=(n+r)!n!\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 (n+r)!\left( {n + r} \right)! appear just as frequently or more as in n!r!n!r!, then now for some integer kk that (n+r)!=k×n!×r!\left( {n + r} \right)! = k \times n! \times r!.
So, we have (n+r)!n!=k×n!×r!n!=k×r!\dfrac{{\left( {n + r} \right)!}}{{n!}} = \dfrac{{k \times n! \times r!}}{{n!}} = k \times r!
Hence, the product of rr consecutive natural numbers are (n+r)(n+r−1)............................................(n+1)=k×r!\left( {n + r} \right)\left( {n + r - 1} \right)............................................\left( {n + 1} \right) = k \times r!
Clearly, the product of rr consecutive natural numbers are divisible by r!r! as it is a factor of the product of the rr 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,............1,2,3,............. To solve these kinds of problems always remember the formula n+rPr=(n+r)!(n+r−r)!=(n+r)!n!=(n+r)(n+r−1).........................(n+1){}^{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).