SATHEE: Permutations And Combinations Question 15

Permutations And Combinations Question 15

Question: The value of $ 2^{n}{1.3.5…..(2n-3)(2n-1)} $ is

Options:

A) $ \frac{(2n)!}{n!} $

B) $ \frac{(2n)!}{2^{n}} $

C) $ {2^{n-i}} $

D) None of these

Show Answer

Answer:

Correct Answer: A

Solution:

$ 1\ .\ 3\ .\ 5……(2n-1)2^{n} $ $ =\frac{1.2.3.4.5.6….(2n-1)(2n)2^{n}}{2.4.6…..2n} $ $ =\frac{(2n)!\ 2^{n}}{2^{n}(1\ .\ 2\ .\ 3……n)}=\frac{(2n)\ !}{n\ !} $ .

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Permutations And Combinations Question 15

Question: The value of $ 2^{n}{1.3.5…..(2n-3)(2n-1)} $ is

Options:

Answer:

Solution:

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