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

$r!$

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$r^{2}$

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$r^{n}$

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none of these

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Solution

The correct option is A
$r!$

Explanation for the correct option:
Product of $r$ consecutive terms $= n (n - 1) (n - 2) . . . . (n - r + 1)$
We know $^{n} C_{r} = n a t u r a l n u m b e r$
$\begin{array}{rcl} ^{n} C_{r} & = & \frac{n!}{r! (n - r)!} = \frac{n (n - 1) (n - 2) (n - 3) . . . . (n - r + 1) . . . 3.2.1}{r! (n - r) (n - r - 1) (n - r - 2) . . . . . . . 3.2.1} \\ = & \frac{n (n - 1) (n - 2) (n - 3) . . . . (n - r + 1) (n - r) (n - r - 1) (n - r - 2) . . . 3.2.1}{r! (n - r) (n - r - 1) (n - r - 2) . . . . . . . 3.2.1} \\ = & \frac{n (n - 1) (n - 2) (n - 3) . . . . (n - r + 1)}{r!} \\ = & \frac{P r o d u c t o f r c o n s e c u t i v e i n t e r g e r}{r!} = n a t u r a l n u m b e r \end{array}$
Hence we can say product of any $r$ consecutive natural numbers is always divisible by $r!$
Therefore the product of $r$ consecutive natural numbers is divisible by the factorial of $r$
Hence, option A is correct

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