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

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) \dots (p - n + 1)

\frac{P}{n!} = \frac{p (p - 1) \dots (p - n + 1)}{n!}

= \frac{[p (p - 1) \dots (p - n + 1)] [(p - n) (p - n - 1) \dots 1]}{n!}

= \frac{p!}{n! (p - n)!} =^{p} C_{n}

^{p} C_{n} is always an integer so P is divisible by n! .