(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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(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:$P = \frac{(n+r)!}{n! r!} \times r! = \binom{n+r}{r} \times r!$.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 $r!$.

Class 11 Mathematics Permutations and Combinations Definition of combinations, Condition combinations

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

WBJEE 2009 All WBJEE

A
$r!$
B
$(r+4)!$
C
$(r+1)!$
D
$(r+2)!$

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Class 11

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Permutations and Combinations

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Definition of combinations, Condition combinations

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

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