A) \[n>6\]
B) \[n>7\]
C) \[n<6\]
D) None of these
Correct Answer: A
Solution :
\[{}^{n}{{C}_{3}}+{}^{n}{{C}_{4}}>{}^{n+1}{{C}_{3}}\] Þ \[{}^{n+1}{{C}_{4}}>{}^{n+1}{{C}_{3}}\,\,(\because \,{}^{n}{{C}_{r}}+{}^{n}{{C}_{r+1}}={}^{n+1}{{C}_{r+1}})\] Þ \[\frac{{}^{n+1}{{C}_{4}}}{{}^{n+1}{{C}_{3}}}>1\]Þ \[\frac{n-2}{4}>1\] \[\Rightarrow n>6\].
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