MCQ
If $^n{C_3} + {\,^n}{C_4} > {\,^{n + 1}}{C_3},$ then
  • $n > 6$
  • B
    $n > 7$
  • C
    $n < 6$
  • D
    None of these

Answer

Correct option: A.
$n > 6$
a
(a) ${}^n{C_3} + {}^n{C_4} > {}^{n + 1}{C_3}$

==>${}^{n + 1}{C_4} > {}^{n + 1}{C_3}\,\,(\,{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}})$

==> $\frac{{{}^{n + 1}{C_4}}}{{{}^{n + 1}{C_3}}} > 1$

==> $\frac{{n - 2}}{4} > 1$

$ \Rightarrow n > 6$.

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