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Question: Prove that \(\dfrac{\left( 2n \right)!}{n!}=1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n...

Prove that (2n)!n!=1⋅3⋅5⋅...⋅(2n−1)2n.\dfrac{\left( 2n \right)!}{n!}=1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}.

Explanation

Solution

The factorial of a natural number is defined as the product of all the natural numbers less than or equal to that number. That is the factorial of a natural number nn is the product of the natural numbers from 11 to n.n. So, n!=1⋅2⋅3⋅...⋅(n−1)n.n!=1\cdot 2\cdot 3\cdot ...\cdot \left( n-1 \right)n.

Complete step by step answer:
Consider the given problem (2n)!n!=1⋅3⋅5⋅...⋅(2n−1)2n.\dfrac{\left( 2n \right)!}{n!}=1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}.
We have already learnt that the factorial of a natural number is defined as the product of all the natural numbers less than or equal to that number.
Suppose we need to find the factorial of a natural number n.n. So, we will get the identity n!=1⋅2⋅3⋅...⋅(n−1)n.n!=1\cdot 2\cdot 3\cdot ...\cdot \left( n-1 \right)n.
Let us consider the right-hand side of the given equation, 1⋅3⋅5⋅...⋅(2n−1)2n.1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}.
Here, the product of the odd numbers from 11 to 2n−12n-1 is multiplied to 2n.{{2}^{n}}.
Now, we need to make this a fraction in terms of factorials. In order to make this in terms of factorial we are going to multiply and divide it with the product 2⋅4⋅6⋅...⋅2n.2\cdot 4\cdot 6\cdot ...\cdot 2n.
So, we will get 1⋅3⋅5⋅...⋅(2n−1)2n=1⋅3⋅5⋅...⋅(2n−1)2n2⋅4⋅6⋅...⋅2n2⋅4⋅6⋅...⋅2n.1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}=1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}\dfrac{2\cdot 4\cdot 6\cdot ...\cdot 2n}{2\cdot 4\cdot 6\cdot ...\cdot 2n}.
We know that 1⋅3⋅5⋅...⋅(2n−1)2n⋅2⋅4⋅...⋅2n=1⋅2⋅3⋅4⋅5⋅...⋅2n⋅2n.1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}\cdot 2\cdot 4\cdot ...\cdot 2n=1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot 2n\cdot {{2}^{n}}.
So, we will get 1⋅3⋅5⋅...⋅(2n−1)2n=1⋅2⋅3⋅4⋅5⋅...⋅2n⋅2n2⋅4⋅...⋅2n.1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}=\dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot 2n\cdot {{2}^{n}}}{2\cdot 4\cdot ...\cdot 2n}.
By the definition of the factorial, we will get the following 1⋅2⋅3⋅4⋅5⋅...⋅2n=2n!.1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot 2n=2n!.
Therefore, we will get the numerator of the above fraction in terms of factorial.
That is, 1⋅2⋅3⋅4⋅5⋅...⋅2n⋅2n=(2n)!2n2⋅4⋅...⋅2n.1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot 2n\cdot {{2}^{n}}=\dfrac{\left( 2n \right)!{{2}^{n}}}{2\cdot 4\cdot ...\cdot 2n}.
Let us consider the denominator of the above fraction, 2⋅4⋅...⋅2n.2\cdot 4\cdot ...\cdot 2n. Since all the terms are even, all of them have a common factor. That is, 2.2. If we are taking 22 from each of these numbers, we will get 2⋅4⋅6⋅...⋅2n=2⋅2⋅2⋅...⋅2ntimes⋅1⋅2⋅3⋅...⋅n.2\cdot 4\cdot 6\cdot ...\cdot 2n=2\cdot 2\cdot 2\cdot ...\cdot {{2}_{n times}}\cdot 1\cdot 2\cdot 3\cdot ...\cdot n.
Thus, we will get 1⋅2⋅3⋅4⋅5⋅...⋅2n⋅2n=(2n)!2n1⋅2⋅3⋅4⋅5⋅...⋅n⋅2n.1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot 2n\cdot {{2}^{n}}=\dfrac{\left( 2n \right)!{{2}^{n}}}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot n\cdot {{2}^{n}}}.
Let us cancel the common factor from the numerator and the denominator to get 1⋅2⋅3⋅4⋅5⋅...⋅2n⋅2n=(2n)!1⋅2⋅3⋅4⋅5⋅...⋅n.1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot 2n\cdot {{2}^{n}}=\dfrac{\left( 2n \right)!}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot n}.
We have 1⋅2⋅3⋅4⋅5⋅...⋅n=n!.1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot n=n!.
Therefore, 1⋅2⋅3⋅4⋅5⋅...⋅2n⋅2n=(2n)!n!=LHS.1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot 2n\cdot {{2}^{n}}=\dfrac{\left( 2n \right)!}{n!}=\text{LHS}.
Hence it is proved that (2n)!n!=1⋅3⋅5⋅...⋅(2n−1)2n.\dfrac{\left( 2n \right)!}{n!}=1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}.

Note: Remember that 0!=1.0!=1. Also, note that (2n)!≠2n!.\left( 2n \right)!\ne 2n!. We know that if we multiply a number multiple times, then that is equal to the number raised to the number of the times it is multiplied.