The product of any r consecutive natural numb | vedclass

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Question

(A) Let the $r$ consecutive natural numbers be $(n+1), (n+2), \dots, (n+r)$.Their product is $P = (n+1)(n+2) \dots (n+r)$.We can write this as:$P = \f

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$r!$

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(A) Let the $r$ consecutive natural numbers be $(n+1), (n+2), \dots, (n+r)$.Their product is $P = (n+1)(n+2) \dots (n+r)$.We can write this as:$P = \frac{(n+r)!}{n!}$.Multiplying and dividing by $r!$,we get:$P = \frac{(n+r)!}{n! r!} 	imes r! = \binom{n+r}{r} 	imes r!$.Since $\binom{n+r}{r}$ is the number of ways to choose $r$ objects from $n+r$ objects,it is always an integer.Therefore,the product $P$ is always divisible by $r!$.

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

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