,
Ashish Menon,
Pi Han Goh, and
contributed
For any non-negative integer \(n,\) we find that
\[\dfrac{n!}{n!!}=(n-1)!! ~~\text{ or }~~ n!=(n-1)!!×n!!.\]
We have the following 2 cases:
- If \(n\) is odd, \[\dfrac{n!}{n!!}=\dfrac{n\times (n-1)\times (n-2)\times \cdots \times 3\times 2\times 1}{n\times (n-2)\times (n-4)\times \cdots 5\times 3\times 1}.\] Since all the odd numbers \(n, n-2, n-4, \ldots , 5, 3\) get canceled, we are left with the equation \[\dfrac{n!}{n!!}=(n-1)!!.\]
- If \(n\) is even, \[\dfrac{n!}{n!!}=\dfrac{n\times (n-1)\times (n-2)\times \cdots \times 3\times 2\times 1}{n\times (n-2)\times (n-4)\times \cdots \times 4\times 2}.\] Since all the even numbers \(n, n-2, n-4, \ldots , 4, 2\) get canceled, we are left with the equation \[\dfrac{n!}{n!!}=(n-1)!!.\]
Combining both cases, we find that for any non-negative integer \(n\), \[\dfrac{n!}{n!!}=(n-1)!!. \ _\square\]
Suppose that \(n!!\) is defined as follows:
\[ n!! = \begin{cases} n \times (n-2) \times \cdots \times 5 \times 3 \times 1 &\text{if } n \text{ is odd}; \\ n \times (n-2) \times \cdots \times 6 \times 4 \times 2 &\text{if } n \text{ is even}; \\ 1 &\text{if } n = 0, - 1. \\ \end{cases} \]
Then what is
\[\color{red}{\dfrac{9!}{6!!}} \div \color{green}{\dfrac{9!!}{6!}}?\]
The correct answer is: 5760
For any non-negative integer \(n,\) we find that
\[\dfrac{(2n+1)!}{(2n)!!}=(2n+1)!!.\]
Here, there is no need to consider two separate cases because it makes no difference whether \(n\) is odd or even.
We can expand the LHS as
\[\dfrac{(2n+1)\times (2n)\times (2n-1)\times \cdots \times 3\times 2\times 1}{(2n)\times (2n-2)\times (2n-4)\times \cdots \times 4\times 2}.\]
Since all the even numbers \(2n, 2n-2, 2n-4, \ldots, 4, 2\) get canceled, we are left with the equation
\[\dfrac{(2n+1)!}{(2n)!!}=(2n+1)!!. \ _\square\]
Evaluate \(\frac {9!}{9!!}\).
Since \(\frac {n!}{n!!}=(n-1)!!\), substituting the values, we get
\[\begin{align} \dfrac{9!}{9!!}&=(9-1)!!\\ &=8!!\\ &=8×6×4×2\\ &=384. \ _\square \end{align}\]
Evaluate \(\frac {(3!)!}{3!!}\).
We have
\[\begin{align} \dfrac {(3!)!}{3!!} &=\dfrac {(3×2×1)!}{3×1}\\ &=\dfrac {6!}{3}\\ &=\dfrac {6×5×4×3×2×1}{3}\\ &=\dfrac {720}{3}\\ &=240. \ _\square \end{align}\]
\[\Large{\color{green}{\dfrac{9!}{8!!}}} \div {\color{orange}{\dfrac{7!}{6!!}}} = \, ? \]
Notation:
\[ n!! = \begin{cases} n \times (n-2) \times \cdots \times 5 \times 3 \times 1 && \text{if } n \text{ is odd;} \\ n \times (n-2) \times \cdots \times 6 \times 4 \times 2 && \text{if } n \text{ is even;} \\ 1 && \text{if } n = 0, - 1. \\ \end{cases} \]
Try the first part here!The correct answer is: 9
For any non-negative integer \(n\) we find that
\[\dfrac{(2n-1)!}{(2n-2)!!}=(2n-1)!!.\]
Again here, there is no need to consider two separate cases. We can expand the LHS as
\[\dfrac{(2n-1)\times (2n-2)\times (2n-3)\times \cdots \times 3\times 2\times 1}{(2n-2)\times (2n-4)\times \cdots \times 4\times 2}.\]
Since all the even numbers \(2n-2, 2n-4, \ldots , 4, 2\) get canceled, we are left with the equation
\[\dfrac{(2n-1)!}{(2n-2)!!}=(2n-1)!!. \ _\square\]
Evaluate \(\frac{9!}{8!!}\).
We have
\[ \begin{align}
\dfrac { 9! } { 8!!} &= \dfrac{ (2 \times 5 - 1 ) ! } { (2 \times 5 - 2 ) !! } \\ &= (2 \times 5 - 1 )!! \\ &= 9 !! \\ &= 9 \times 7 \times 5 \times 3 \times 1 \\ &= 945. \ _\square \end{align} \]
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