MCQ
If $^n{C_3} + {\,^n}{C_4} > {\,^{n + 1}}{C_3},$ then
- ✓$n > 6$
- 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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