Permutations And Combinations Question 362
Question: Let $ E=(2n+1)(2n+3)(2n+5)…(4n-3)(4n-1); $ $ n>1 $ then $ 2^{n}E $ divisible by
Options:
A) $ ^{n}{C _{n/2}} $
B) $ ^{2n}C _{n} $
C) $ ^{3n}C _{n} $
D) $ ^{4n}C _{2n} $
Correct Answer: D $ \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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Answer:
Solution:
BETA
