A) 7
B) 8
C) 9
D) 10
Correct Answer: B
Solution :
[b] We have \[^{10}{{C}_{n-1}}>{{2.}^{10}}{{C}_{n}}\] \[\Rightarrow \frac{10!}{(n-1)!(11-n)!}>2.\frac{10!}{n!(10-n)!}\] \[\Rightarrow \frac{n!}{(n-1)!}>2\frac{(11-n)!}{(10-n)!}\] i.e, \[n>22-2n\] \[\Rightarrow 3n>22\Rightarrow n>\frac{22}{3}=7\frac{1}{3}\] \[\therefore \] Least + ve integral value of n=8.You need to login to perform this action.
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