The product of any r consecutive natural numb | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$r!$

Vedclass · Answer

(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

Similar Questions

If $\binom{n}{3} + \binom{n}{4} > \binom{n+1}{3}$,then:

If ${ }^{15} C_4+{ }^{15} C_5+{ }^{16} C_6+{ }^{17} C_7+{ }^{18} C_8={ }^{19} C_{r}$,then the value of $r$ is equal to

If $^{2n}C_3 : ^nC_2 = 44:3$,then for which of the following values of $r$,the value of $^nC_r$ will be $15$?

If ${ }^{n} C_{12}={ }^{n} C_{8}$,then $n$ is equal to

$A$ class contains $b$ boys and $g$ girls. If the number of ways of selecting $3$ boys and $2$ girls from the class is $168$,then $b + 3g$ is equal to.

Vedclass Test Series

Exam Paper Generator

Online Exam Module