# What would be the expansion of (2n)! , 2n! , (2n)!! and 2n!! ? Kindly comprehend the answer in detail.

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### 1 Answer

Hello!

The factorial of natural m is defined as a product of all natural numbers less than or equal to m, i.e. m! = 1*2*...*(m-1)*m. Also 0! is defined as 1.

Therefore, (2n)! = 1*2*...*(2n-1)*(2n). For example, (2*0)! = 1, (2*1)! = 1*2 = 2, (2*2)! = 1*2*3*4 = 24. (2*3)! = 1*2*3*4*5*6 = 720.

2n! which I understand as 2*(n!) is equal to 2*1*2*...*(n-1)*n and is almost always less then (2n)!:

2*0! = 2, 2*1! = 2, 2*2! = 4, 2*3! = 12.

Now for (2n)!!. By definition, m!! is the product of all natural numbers **with the same parity as m** less than or equal to m:

m!! = m*(m-2)*(m-4)*...*(2 or 1).

and 0!! = 1

For even m there are m/2 factors and the last factor is 2. For odd m there are (m+1)/2 factors and the last factor is 1.

So, (2n)!! = 2*4*...*(2n-2)*(2n).

0!! = 1, 2!! = 2, 4!! = 2*4 = 8, 6!! = 2*4*6 = 48, 8!! = 2*4*6*8 = 384.

2*n!! is a different thing: 2*n*(n-2)*...*(2 or 1).

2*0!! = 2, 2*1!! = 2, 2*2!! = 4, 2*3!! = 6, 2*4!! = 16.

I hope this answers your question.

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