A) 11
B) 10
C) 12
D) 13
Correct Answer: A
Solution :
[a] : \[^{n}{{C}_{5}}{{+}^{n}}{{C}_{6}}{{>}^{n+1}}{{C}_{5}}{{\Rightarrow }^{n+1}}{{C}_{6}}{{>}^{n+1}}{{C}_{5}}\] \[\Rightarrow \]\[\frac{(n+1)!}{6!(n-5)!}.\frac{5!(n-4)!}{(n+1)!}>1\Rightarrow \frac{(n-4)}{6}>1\] \[\Rightarrow \]\[n-4>6\Rightarrow n>10\] Hence, least value of n = 11.
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