Prove that product of any n consecutive integers is divisiblr by n!


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Question Number 769 by rishabh last updated on 09/Mar/15

$P r o v e t h a t p r o d u c t o f a n y n c o n s e c u t i v e i n t e g e r s$ $i s d i v i s i b l r b y n!$
Commented by 123456 last updated on 09/Mar/15

$n (n + 1) \cdot \cdot \cdot (2 n - 1)$
	Answered by prakash jain last updated on 09/Mar/15

	$n consecutive number ending with p are (P)$ $P = p (p - 1) . . . (p - n + 1)$ $\frac{P}{n!} = \frac{p (p - 1) . . . (p - n + 1)}{n!}$ $= \frac{[p (p - 1) . . . (p - n + 1)] [(p - n) (p - n - 1) . . . 1]}{n!}$ $= \frac{p!}{n! (p - n)!} =^{p} C_{n}$ $^{p} C_{n} is always an integer so P is divisible by n! .$
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