JEE Main & Advanced Mathematics Permutations and Combinations Question Bank Self Evaluation Test - Permutations and Combinatioins

  • question_answer
    Let \[E=(2n+1)(2n+3)(2n+5)...(4n-3)(4n-1);\] \[n>1\] then \[{{2}^{n}}E\] divisible by

    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 \[^{4n}{{C}_{2n}}.\]

You need to login to perform this action.
You will be redirected in 3 sec spinner