Factorial fun! Part-2

Now using the property $\boxed{\dfrac{n!}{(n-1)!!}=n!!}$

• we find that: $\dfrac{9!}{(9-1)!!}=9!! \\ =9×(9-2)×(9-4)×(9-6) \\ =9×7×5×3 \\ =\boxed{945}$

• Similarly, $\dfrac{7!}{(7-1)!!}=7!! \\ =7×(7-2)×(7-4) \\ =7×5×3 \\ =\boxed{105}$

Finally, we need to find: $\dfrac{945}{105}=\boxed{9}$

We can find this using, another property which I'd observed i.e. $\boxed{\dfrac{(2n+1)!}{(2n)!!}=(2n+1)!!}$

$\frac{9! \times 6!!}{7! \times 8!!}$

$\frac{9 \times 8 \times 7! \times 6!!}{7! \times 8 \times 6!!}$

$\boxed 9$

$\displaystyle \frac { 9! } { 8!! } / \frac { 7! } { 6!! } = \frac { 9! } { 8!! } \times \frac { 6!! } { 7! } = \frac { 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 6 \times 4 \times 2 } { 8 \times 6 \times 4 \times 2 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 } = \boxed { 9 }$ .

$\boxed { \text { The answer is \boxed { 9 } } }$ .