`""^(2n)P_(n)` is equal to

Options

A. `(n+1)(n+2)…(2n)`
B. `2^(n)[1*3*5….(2n-1)]`
C. `(2)*(6)*(10)…(4n-2)`
D. `n!(""^(2n)C_(n))`

Answer

To solve the problem of finding \( (2n)P_n \), we will use the formula for permutations. The formula for \( nPr \) is given by:

\[ nPr = \frac{n!}{(n-r)!} \]

In our case, we need to find \( (2n)P_n \):

\[ (2n)P_n = \frac{(2n)!}{(2n-n)!} = \frac{(2n)!}{n!} \]

Now, we can express \( (2n)! \) in terms of its factors:

\[ (2n)! = 1 \times 2 \times 3 \times \ldots \times (2n) \]

We can separate the even and odd factors in \( (2n)! \):

1. **Odd factors**: The odd numbers from 1 to \( 2n \) are \( 1, 3, 5, \ldots, (2n-1) \).
2. **Even factors**: The even numbers from 1 to \( 2n \) are \( 2, 4, 6, \ldots, 2n \).

The product of the odd numbers can be expressed as:

\[ 1 \times 3 \times 5 \times \ldots \times (2n-1) = \frac{(2n)!}{2^n \cdot n!} \]

The product of the even numbers can be expressed as:

\[ 2 \times 4 \times 6 \times \ldots \times (2n) = 2^n \times (1 \times 2 \times 3 \times \ldots \times n) = 2^n \cdot n! \]

Thus, we can write \( (2n)! \) as:

\[ (2n)! = (1 \times 3 \times 5 \times \ldots \times (2n-1)) \times (2 \times 4 \times 6 \times \ldots \times (2n)) \]

Now substituting back into our permutation formula:

\[ (2n)P_n = \frac{(2n)!}{n!} = \frac{(1 \times 3 \times 5 \times \ldots \times (2n-1)) \times (2^n \cdot n!)}{n!} \]

The \( n! \) cancels out:

\[ (2n)P_n = (1 \times 3 \times 5 \times \ldots \times (2n-1)) \times 2^n \]

Now, we can also express \( (2n)P_n \) in another way by recognizing that:

\[ (2n)P_n = n! \cdot (n+1)(n+2)\ldots(2n) \]

This gives us:

\[ (2n)P_n = n! \cdot \frac{(2n)!}{n!} = (n+1)(n+2)\ldots(2n) \]

Thus, we conclude that:

\[ (2n)P_n = n! \cdot (n+1)(n+2)\ldots(2n) \]

### Summary of Results:
1. \( (2n)P_n = \frac{(2n)!}{n!} \)
2. \( (2n)P_n = (1 \times 3 \times 5 \times \ldots \times (2n-1)) \times 2^n \)
3. \( (2n)P_n = (n+1)(n+2)\ldots(2n) \)