Question

The product of n consecutive natural numbers is always divisible by:
A) 4n!
B) 3n!
C) 2n!
D) n!

Answer 1

Hint: We will assume the n consecutive numbers with the common assumption. Then we will rewrite them using some modifications and thus get the required answer.

Complete step-by-step solution:
We will solve this question starting with taking assumption numbers.
Let the n consecutive numbers to be k + 1, k + 2, ………. And the last one is k + n.
Now, let us see their product.
We will get: (k + 1)(k + 2)(k + 3)………..(k + n).
We can rewrite this product by multiplying and dividing the same by k! to get the following expression:-
$ \Rightarrow $ Product = $\dfrac{{k!(k + 1)(k + 2)........(k + n)}}{{k!}}$
Since we know that $n! = 1.2.3.......(n - 1).n$
Therefore, we will obtain:-
$ \Rightarrow $ Product = $\dfrac{{(k + n)!}}{{k!}}$
Let us now multiply and divide this by n! in order to get:-
$ \Rightarrow $ Product = $\dfrac{{n! \times (k + n)!}}{{n! \times k!}}$
Since we know that $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$.
So, we will get:-
$ \Rightarrow $ Product = $n!\left( {^{n + k}{C_n}} \right)$
Since we can clearly see that it has a factor of n! in it. Therefore, it must be definitely divisible by n!

$\therefore $ The correct option is (D).

Note: The students must commit to memory the following formulas:
$n! = 1.2.3.......(n - 1).n$
$^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
The students must keep in mind that they cannot just take n numbers as 1, 2 , 3 and so on up to n because that will be the first n consecutive natural numbers but in the question, we are not given any particular sequence, therefore, we must take arbitrary numbers. Taking k in addition makes them arbitrary because we can take any random value of k and thus get the required sequence of n consecutive natural numbers.