MCQ
The product of $r$ consecutive positive integers is divisible by:
- ✓$r!$
- B$(r − 1)!$
- C$(r + 1)!$
- DNone of these
✓
Answer
Correct option: A.
$r!$
The product of $r$ consecutive integers is equal to $r!,$
so it will be divisible by $r!.$
so it will be divisible by $r!.$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The diagram shows the graph of $y = a{x^2} + bx + c$. Then
→If $\sin \theta + \cos \theta = \sqrt 2 \cos \alpha $, then the general value of $\theta $ is
→Let $z$ be a complex number. Then the angle between vectors $z$ and $ - iz$ is
→A line passes through the point of intersection of $2x + y = 5$ and $x + 3y + 8 = 0$ and parallel to the line $3x + 4y = 7$ is
→If $\alpha$ denotes the number of solutions of $|1-i|^x=2^x$ and $\beta=\left(\frac{|z|}{\arg (z)}\right)$, where $z=\frac{\pi}{4}(1+i)^4\left(\frac{1-\sqrt{\pi i}}{\sqrt{\pi}+i}+\frac{\sqrt{\pi}-i}{1+\sqrt{\pi} \mathrm{i}}\right), i=\sqrt{-1}$, then the distance of the point $(\alpha, \beta)$ from the line $4 x-3 y=7$ is
→If $x, y, z \in R^+$ such that $x + y + z = 4$, then maximum possible value of $xyz^2$ is -
→For what value of $\theta, 1$ lies between the roots of the quadratic equation $3\text{x}^2 – 3\sin\theta \text{x} – 2\cos^2\theta = 0$
→Let
→$A=\left\{(x, y) \in R \times R \mid 2 x^{2}+2 y^{2}-2 x-2 y=1\right\}$
$B=\left\{(x, y) \in R \times R \mid 4 x^{2}+4 y^{2}-16 y+7=0\right\} \text { and }$
$C=\left\{(x, y) \in R \times R \mid x^{2}+y^{2}-4 x-2 y+5 \leq r^{2}\right\}$
Then the minimum value of $|r|$ such that $A \cup B \subseteq C$ is equal to:
If general term of an $A.P.$ is $3n$ then find common difference.
→The locus of the centre of a circle which passes through the point $(a, 0)$ and touches the line $x + 1 = 0$, is
→