- ✓$n > 6$
- B$n > 7$
- C$n < 6$
- DNone of these
==>${}^{n + 1}{C_4} > {}^{n + 1}{C_3}\,\,(\,{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}})$
==> $\frac{{{}^{n + 1}{C_4}}}{{{}^{n + 1}{C_3}}} > 1$
==> $\frac{{n - 2}}{4} > 1$
$ \Rightarrow n > 6$.
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$(A)$ $\left|z-z_1\right|+\left|z-z_2\right|=\left|z_1-z_2\right|$
$(B)$ $\operatorname{Arg}\left(z-z_1\right)=\operatorname{Arg}\left(z-z_2\right)$
$(C)$ $\left|\begin{array}{cc}z-z_1 & \bar{z}-\bar{z}_1 \\ z_2-z_1 & \bar{z}_2-\bar{z}_1\end{array}\right|=0$
$(D)$ $\operatorname{Arg}\left(z-z_1\right)=\operatorname{Arg}\left(z_2-z_1\right)$
$I.$ If $n$ is a composite number, then $n$ divides $(n-1) ! .$
$II$. There are infinitely many natural numbers $n$ such that $n^3+2 n^2+n$ divides $n !$