A) \[^{n}{{C}_{n/2}}\]
B) \[^{2n}{{C}_{n}}\]
C) \[^{3n}{{C}_{n}}\]
D) \[^{4n}{{C}_{2n}}\]
Correct Answer: D
Solution :
[d] Here, \[E=(2n+1)(2n+3)(2n+5)...(4n-3)(4n-1)\] Or \[(2n+1).(2n+2).(2n+3).(2n+4)\] Or \[E=\frac{..............(4n-1)(4n)}{(2n+2)(2n+4)...(4n)}.\frac{(2n)!}{(2n)!}\] \[E=\frac{(4n)!\,\cdot n!}{(2n)!{{2}^{n}}\cdot (2n)!}\] \[\Rightarrow {{2}^{n}}E=\frac{(4n)!}{(2n)!(2n)!}.\,n!{{=}^{4n}}{{C}_{2n}}.n!\] \[\Rightarrow {{2}^{n}}E\] is divisible by \[^{4n}{{C}_{2n}}.\]
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