Question

How do you simplify \[\dfrac{{2n!}}{{n!}}\] ?

Answer 1

Hint: Here in this question, they have used the \[ \div \] operator. This operator does the division and we are dividing the two terms. In this question we also see ! symbol this represents the factorial, the factorial is defined as \[n! = n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1\] . So by the definition of factorial and division we are going to solve this and find the solution.

Complete step-by-step answer:
The factorial is defined as \[n! = n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1\] . The 2n!, 2n factorial is defined as \[2n! = 2n \times (2n - 1) \times (2n - 2) \times ... \times 2 \times 1\] . In the n! have n terms.
Here in this question, we have to simplify \[\dfrac{{2n!}}{{n!}}\] ------ (1)
We know the expansion of 2n! and n! , substituting these expansion in the equation (1) we have
 \[ \Rightarrow \dfrac{{2n \times (2n - 1) \times (2n - 2) \times (2n - 4)...3 \times 2 \times 1}}{{n \times (n - 1) \times (n - 2)...3 \times 2 \times 1!}}\]
In the numerator we can take 2 as a common in some terms like (2n-2), (2n-4) and so on.
By taking the common the above expression or inequality is written as
 \[ \Rightarrow \dfrac{{2n \times (2n - 1) \times 2(n - 1) \times 2(n - 2)...3 \times 2 \times 1}}{{n \times (n - 1) \times (n - 2)...3 \times 2 \times 1!}}\]
If we see both numerator and denominator some terms are the same or similar. So we can cancels the terms and we have
 \[ \Rightarrow 2 \times (2n - 1) \times 2 \times (2n - 3) \times 2...5 \times 3 \times 1\]
By altering the above equation or inequality it is rewritten as
 \[ \Rightarrow \left( {(2n - 1) \times (2n - 3)...5 \times 3 \times 1} \right)\left( {2 \times 2 \times 2 \times ...} \right)\]
When we see the first term of the above equation it represents the sequence of odd numbers. In the second term the number 2 is multiplied n times so we have
 \[ \Rightarrow \dfrac{{2n!}}{{n!}} = 1 \times 3 \times 5 \times ..... \times (2n - 3) \times (2n - 1)({2^n})\]
Therefore we have
 \[\dfrac{{2n!}}{{n!}} = \left( {1 \times 3 \times 5 \times ..... \times (2n - 3) \times (2n - 1)} \right)({2^n})\]
So, the correct answer is “ \[\dfrac{{2n!}}{{n!}} = \left( {1 \times 3 \times 5 \times ..... \times (2n - 3) \times (2n - 1)} \right)({2^n})\] ”.

Note: To solve the mathematical problems we have mathematical operations. There are for mathematical operations namely addition, subtraction, multiplication and division. + represents the addition, - represents subtraction, \[ \times \] represents the multiplication and \[ \div \] represents the division. The multiplication is the repeated addition. The division is repeated subtraction. ! indicates the factorial. In general the factorial is defined as \[n! = n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1\] where n is a positive integer.