Permutations And Combinations Question 365
Question: The least positive integral values of n which satisfies the inequality $ ^{10}{C _{n-1}}>{{2.}^{10}}C _{n} $
Options:
A) 7
B) 8
C) 9
D) 10
Correct Answer: B $ \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.
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Answer:
Solution:
BETA
