Question

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Question : 1 × 3 × 5 × 7 × 9 × ......... × ( 2n - 1 ) =


$1. \: \mathsf{\dfrac{(2n)!}{ {2}^{n}.n!}}$


$2. \: \mathsf{\dfrac{(3n)!}{ {2}^{n}.n!}}$


$3. \: \mathsf{(3n)!}$


CHAPTER : PERMUTATIONS AND COMBINATION



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Answer 1

Answer:

1

Step-by-step explanation:

Given: 1 * 3 * 5 .... (2n - 1).

Multiply and Divide by 2,4,6...2n, we get

$=1.3.5.....(2n - 1) * \frac{2.4.6...2n}{2.4.6...2n}$

$=\frac{1.2.3.4.5....(2n-1)(2n)}{2.4.6....2n}$

$=\frac{(2n)(2n-1)...5.4.3.2.1}{2.4.6...2n}$

$=\frac{2n!}{2.4.6...2n}$

$=\frac{2n!}{(2*1)(2*2)...(2*n)}$

$=\frac{2n!}{2^n * (1.2.3...n)}$

$=\boxed{\frac{2n!}{2^n * n!}}$

Hope it helps!

Answer 2

2nCn =  (2n)! / n! n!

= (2n)! / (n!)2

= (2n) ( 2n - 1) (2n - 2) -----------4.3.2.1 / (n!)2

= 2xn ( 2n - 1) 2 ( n - 1) -----------------2x2.3.2x1.1 / (n!)2

= 2n [n(n - 1)(n - 2)(n -3) ----------3 x2 x1][(2n -1)(2n -3) ---------------5x3x1] / (n!)2

= 2n n! [(2n -1)(2n -3) ---------------5x3x1] / (n!)2

= 2n  [(2n -1)(2n -3) ---------------5x3x1] / (n!)

= 2n[ 1x3x5 -----------------(2n -3)(2n -1)] / (n!)