Question
Question: How do you simplify \(\dfrac{(2n)!}{n!}\)?...
How do you simplify n!(2n)!?
Solution
A factorial of a number multiplied by all of the integers preceding it. Example: “Five Factorial” = 5!=5×4×3×2×1=120. Hence factorial of a number n is the product of all the numbers from 1 to n. So we have
n!=n⋅(n−1)⋅(n−2)⋅...⋅2⋅1
Although there is no simplification of n!(2n)!, but there are other ways of expressing it, such as
n!(2n)!=k=0∏n−1(2n−k)=(2n)(2n−1)...(n+1) .
Complete step by step solution:
To simplify: n!(2n)!
By the definition of factorial function,
(2n)!=(2n)(2n−1)...2⋅1
And n!=n(n−1)(n−2)...⋅2⋅1
Now putting the value of (2n)! in n!(2n)! , we get
n!(2n)!=n!1(1⋅2⋅3⋅...⋅2n)
This can also be written as:
n!(2n)!=n!1(2⋅4⋅6⋅...⋅2n)(1⋅3⋅5⋅...⋅(2n−1))=n!1(1⋅2)(2⋅2)(3⋅2)...(n⋅2)(1⋅3⋅5⋅...(2n−1))=n!1(1⋅2⋅3⋅...n)2n(1⋅3⋅5⋅...⋅(2n−1))=n!1n!⋅2n(1⋅3⋅5⋅...⋅(2n−1))=2n(1⋅3⋅5⋅...⋅(2n−1))
Therefore, n!(2n)!=2n(1⋅3⋅5⋅...⋅(2n−1)).
Note: Once we know the value of n which can be either positive, negative, or zero, we can easily find the value of n!(2n)! by putting n in the solution. Factorials have a usage in various areas of mathematics such as probability, combinations and permutations, Taylor’s series, exponential functions, Binomial expansion, and many more.