Permutations And Combinations Question 323
Question: The value of ?n? for which $ ^{n-1}C_4{{-}^{n-1}}C_3-\frac{5}{4}{{.}^{n-1}}P_2<0, $ Where $ n\in N $
Options:
A) $ {5,6,7,8,9,10} $
B) $ {1,2,3,4,5,6,7,8,9,10} $
C) $ {1,4,5,6,7,8,9,10} $
D) $ (-\infty ,2)\cup (3,11) $
Correct Answer: A $ \Rightarrow \frac{(n-1)(n-2)(n-3)(n-4)}{4!}-\frac{(n-1)(n-2)(n-3)}{3!} $ $ -\frac{5}{4}(n-2)(n-3)<0 $ $ \Rightarrow (n-2)(n-3)(n-11)(n+2)<0 $ $ \Rightarrow (n-2)(n-3)(n-11)<0 $ $ [\because n+2>0forn\in N] $ $ \Rightarrow n\in (-\infty ,2)\cup (3,11) $ $ \Rightarrow n\in (0,2)\cup (3,11) $ $ \Rightarrow n=1,4,5,6,7,8,9,10 $
But $ ^{n-1}C_4 $ and $ ^{n-2}P_2 $ both are meaningful for $ n\ge 5. $
Hence,
$ n=5,6,7,8,9,10. $
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Answer:
Solution:
BETA
