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
Solution :
[a] we have, \[^{n-1}{{C}_{4}}{{-}^{n-1}}{{C}_{3}}-\frac{5}{4}{{,}^{n-2}}{{P}_{2}}<0\] |
\[\Rightarrow \frac{(n-1)(n-2)(n-3)(n-4)}{4!}-\frac{(n-1)(n-2)(n-3)}{3!}\] |
\[-\frac{5}{4}(n-2)(n-3)<0\] |
\[\Rightarrow (n-2)(n-3)(n-11)(n+2)<0\] |
\[\Rightarrow (n-2)(n-3)(n-11)<0\] |
\[[\because n+2>0forn\in N]\] |
\[\Rightarrow n\in (-\infty ,2)\cup (3,11)\] |
\[\Rightarrow n\in (0,2)\cup (3,11)\] |
\[\Rightarrow n=1,4,5,6,7,8,9,10\] |
But \[^{n-1}{{C}_{4}}\] and \[^{n-2}{{P}_{2}}\] both are meaningful for \[n\ge 5.\] |
Hence, |
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