Prove that 2 n C n -2n-1-3 -5 -2 n -1- n -

^2nC_n=(2n) !/(n !)(2n 1) !=(2n) !/(n !)^2 =(2n)(2n 1)(2n 2)...4.3.2.1/(n !)^2 =[(2n)(2n 2)(2n 4)...4.2 ]×[(2n 1)(2n 3)...5.3.1 ]/(n !)^2 =2^n[n(n 1)(n 2)...2 ...

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Byju's Answer

Standard XII

Mathematics

nCr Definitions and Properties

Prove that 2 ...

Question

Prove that $^{2 n} C_{n} = \frac{2^{n} \times [1.3.5... (2 n - 1)]}{n!} .$

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Solution

$^{2 n} C_{n} = \frac{(2 n)!}{(n!) (2 n - 1)!} = \frac{(2 n)!}{(n!)^{2}}$

$= \frac{(2 n) (2 n - 1) (2 n - 2) . . .4 .3 .2 .1}{(n!)^{2}}$

$= \frac{[(2 n) (2 n - 2) (2 n - 4) . . .4 .2] \times [(2 n - 1) (2 n - 3) . . .5 .3 .1]}{(n!)^{2}}$

$= \frac{2^{n} [n (n - 1) (n - 2) . . .2 .1] \times [(2 n - 1) (2 n - 3) . . .5 .3 .1]}{(n!)^{2}}$

$= \frac{2^{n} \times (n!) \times [1.3.5... (2 n - 3) (2 n - 1)]}{n!}$

$= \frac{2^{n} \times [1.3.5... (2 n - 3) (2 n - 1)]}{(n!)} .$

Hence, $^{2 n} C_{n} = \frac{2^{n} \times {1 \times 3 \times 5 \times . . . \times (2 n - 1)}}{n!} .$

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Prove that 2nCn=2n×[1.3.5...(2n−1)]n !.

Prove that $^{2 n} C_{n} = \frac{2^{n} \times [1.3.5... (2 n - 1)]}{n!} .$