The product of any r consecutive natural numb | vedclass

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(A) Let the $r$ consecutive natural numbers be $n, n+1, n+2, \dots, n+r-1$. Their product is given by $P = n(n+1)(n+2)\dots(n+r-1)$. We know that the

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$r!$

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(A) Let the $r$ consecutive natural numbers be $n, n+1, n+2, \dots, n+r-1$. Their product is given by $P = n(n+1)(n+2)\dots(n+r-1)$. We know that the number of ways to choose $r$ objects from $n+r-1$ objects is given by the binomial coefficient $\binom{n+r-1}{r} = \frac{(n+r-1)!}{r!(n-1)!}$. This can be rewritten as $\frac{(n+r-1)(n+r-2)\dots(n)}{r!} = \binom{n+r-1}{r}$. Since $\binom{n+r-1}{r}$ is always an integer,it follows that the product $n(n+1)\dots(n+r-1)$ must be divisible by $r!$.

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

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