Factorial fun!

Number Theory Level 2

Suppose that n ! ! n!! is defined as follows:

n ! ! = { n × ( n − 2 ) × ⋯ × 5 × 3 × 1 if n is odd ; n × ( n − 2 ) × ⋯ × 6 × 4 × 2 if n is even ; 1 if n = 0 , − 1. 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

9 ! 6 ! ! ÷ 9 ! ! 6 ! ? \color{#D61F06}{\dfrac{9!}{6!!}} \div \color{#20A900}{\dfrac{9!!}{6!}}?


The answer is 5760.

View wiki
Sravanth C. by Sravanth C. , 21, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

4 solutions

Sravanth C.
Jun 9, 2015

Simplifying the expression we get: 9 ! 9 ! ! × 6 ! 6 ! ! \dfrac{9!}{9!!}×\dfrac{6!}{6!!}

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

  • we find that: 9 ! 9 ! ! = ( 9 − 1 ) ! ! = 8 ! ! = 8 × ( 8 − 2 ) × ( 8 − 4 ) × ( 8 − 6 ) = 8 × 6 × 4 × 2 = 384 \dfrac{9!}{9!!}=(9-1)!! \\ =8!! \\ =8×(8-2)×(8-4)×(8-6) \\ =8×6×4×2 \\ =\boxed{384}

  • Similarly, 6 ! 6 ! ! = ( 6 − 1 ) ! ! = 5 ! ! = 5 × ( 5 − 2 ) × ( 5 − 4 ) = 5 × 3 × 1 = 15 \dfrac{6!}{6!!}=(6-1)!! \\ =5!! \\ =5×(5-2)×(5-4) \\ =5×3×1 \\ =\boxed{15}

Finally, we need to find: 384 × 15 = 5760 384×15=\boxed{5760}

20 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Nicely done! Can you prove the identity n ! n ! ! = ( n − 1 ) ! ! \dfrac{n!}{n!!}=(n-1)!! ?

In response to challenge master @Brilliant Mathematics , sir this is how I wrote the proof in one of the wiki's, can you verify that the proof is correct?

  • Case ( 1 ) (1)

If n n is odd : n ! n ! ! = n × ( n − 1 ) × ( n − 2 ) . . . × 3 × 2 × 1 n × ( n − 2 ) × ( n − 4 ) . . . 5 × 3 × 1 \dfrac{n!}{n!!}=\dfrac{n×(n-1)×(n-2)...×3×2×1}{n×(n-2)×(n-4)...5×3×1}

We find that n n , ( n − 2 ) (n-2) , ( n − 4 ) (n-4) . . . , 5 5 , 3 3 get canceled or, put in other words all the odd numbers get canceled , leading to this equation: n ! n ! ! = ( n − 1 ) ! ! \dfrac{n!}{n!!}=(n-1)!!

  • Case ( 2 ) (2)

If n n is even : n ! n ! ! = n × ( n − 1 ) × ( n − 2 ) . . . × 3 × 2 × 1 n × ( n − 2 ) × ( n − 4 ) . . . 4 × 2 \dfrac{n!}{n!!}=\dfrac{n×(n-1)×(n-2)...×3×2×1}{n×(n-2)×(n-4)...4×2}

We find that n n , ( n − 2 ) (n-2) , ( n − 4 ) (n-4) . . . , 4 4 , 2 2 get canceled or, all the even numbers get canceled , leading us to this: n ! n ! ! = ( n − 1 ) ! ! \dfrac{n!}{n!!}=(n-1)!!

  • Now, if we combine both the cases we find that for any non-negative integer n n : n ! n ! ! = ( n − 1 ) ! ! \boxed{\dfrac{n!}{n!!}=(n-1)!!}

  • Let's see the cases of 0 0 and 1 1 because they may look undefined, but are actually defined.

  • Before this, we must know that ( − 1 ) ! ! = 0 ! ! = 1 (-1)!!=0!!=1

  • First, let n = 0 n=0 :

So, LHS: 0 ! 0 ! ! = 1 1 = 1 \dfrac{0!}{0!!} = \dfrac{1}{1} =1 RHS: ( 0 − 1 ) ! ! ) = ( − 1 ) ! ! = 1 (0-1)!!)=(-1)!!=1

  • Now let n = 1 n=1

So, LHS: 1 ! 1 ! ! = 1 1 = 1 \dfrac{1!}{1!!}= \dfrac{1}{1} =1 RHS: ( 1 − 1 ) ! ! = 0 ! ! = 1 (1-1)!! =0!!=1

Sravanth C. - 6 years ago

Log in to reply

The final conclusion should be "for any positive integer n n " since 0 0 , which is a non-negative integer doesn't satisfy the result.

Also, I think you should show the case of n = 1 n=1 separately since it doesn't quite fit in the proof of the result.

Prasun Biswas - 6 years ago

Log in to reply

Sir, I went for a Google search, and found that − 1 ! ! = 0 ! ! = 1 -1!!=0!!=1 what do you say?

  • I just asked this because, if the above condition is true then my proof would be true. Please reply. . .

Sravanth C. - 6 years ago

Log in to reply

@Sravanth C. – Well, that's new to me. I thought ( − 1 ) ! ! (-1)!! was undefined. Seems like that isn't the case.

Tell you what, you can see that your proof works for all positive integers ≥ 2 \geq 2 without a doubt. It'll be better to mention the case of n = 0 , 1 n=0,1 separately (simply verify the result for those cases, should take one or two lines at max).

Prasun Biswas - 6 years ago

Log in to reply

@Prasun Biswas – I hope it's okay now. ⌣ ¨ \huge\ddot\smile !

Sravanth C. - 6 years ago

Log in to reply

@Sravanth C. – Yes, it's perfect. Upvoted! :)

Prasun Biswas - 6 years ago

Log in to reply

@Prasun Biswas – Thank you very much! :)

Sravanth C. - 6 years ago

Hhi. I missed to times 2 my ans. If i did I would get that correct one! ^_^ X)

Hanifah Fathin Nasution - 6 years ago

Can you explain what is double factorial

Mvs Prakash - 5 years, 11 months ago

Log in to reply

You can read this for more info. Cheers!

Sravanth C. - 5 years, 11 months ago
Ahmed Obaiedallah
Jun 15, 2015

9 ! × 6 ! 9 ! ! × 6 ! ! \frac {9! \times 6!}{9!! \times 6!!}

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

8 × 6 ! = 5760 8 \times 6! =\boxed {\color{#D61F06}{5760}}

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Lew Sterling Jr
Jun 11, 2015

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Gia Hoàng Phạm
Mar 15, 2019

9 ! 6 ! ! ÷ 9 ! ! 6 ! = 9 ! 6 ! ! × 6 ! 9 ! ! = 6 ! × 9 ! 6 ! ! × 9 ! ! = 6 ! 6 ! ! × 9 ! 9 ! ! = ( 6 − 1 ) ! ! × ( 9 − 1 ) ! ! = 5 ! ! × 8 ! ! = 5 × 3 × 1 × 8 × 6 × 4 × 2 = 5760 \begin{aligned} & & \frac{ 9! }{ 6!! } \div \frac{ 9!! }{ 6! } \\ & = \frac{ 9! }{ 6!! } \times \frac{ 6! }{ 9!! } \\ & = \frac{ 6! \times 9! }{ 6!! \times 9!! } \\ & = \frac{ 6! }{ 6!! } \times \frac{ 9! }{ 9!! } \\ & = ( 6 - 1 )!! \times ( 9 - 1 )!! \\ & = 5!! \times 8!! \\ & = 5 \times 3 \times 1 \times 8 \times 6 \times 4 \times 2 \\ & = \boxed{ \large{ 5760 } } \end{aligned}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...