Factorial factorization

Number Theory Level 1

What is the value of 11 ! − 10 ! 9 ! ? \dfrac{11!-10!}{9!}?


Notation: ! ! is the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .

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Jessey Uche-Nwichi by Jessey Uche-Nwichi , 17, Nigeria

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14 solutions

11 ! − 10 ! 9 ! = 10 ! × ( 11 − 1 ) 9 ! = 10 × 10 = 100 \frac{11!-10!}{9!} = \frac{10!\times (11-1)}{9!}=10\times 10 = 100

47 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Factoring is an immensely helpful tool when dealing with factorial expressions where addition or subtraction is involved. Keeping the expansion of factorials in mind (for example, that 10 ! = ( 10 ) ( 9 ) ( 8 ) ( 7 ) ( 6 ) ( 5 ) ( 4 ) ( 3 ) ( 2 ) ) 10! = (10)(9)(8)(7)(6)(5)(4)(3)(2)) will make factoring opportunities easier to spot.

How do you go from 11!-10! = 10!x(11-1) ?

Simon Cochrane - 4 years, 4 months ago

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11! = 11x10x9x...x2=11!x10! Once you know this, you can write the expression as stated above

Djordje Veljkovic - 4 years, 4 months ago

It seems like you only Latex-ed your left bracket but not your right bracket. You can actually group everything into a single Latex.

Christopher Boo - 4 years, 4 months ago

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I'm not very skilled with Latex. The parentheses aren't even supposed to be there, but they show up when I use Latex with fractions.

Djordje Veljkovic - 4 years, 4 months ago

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I think what you want is 

1
 
\frac{11!-10!}{9!} = \frac{10!\times (11-1)}{9!}=10\times 10 = 100

This will render

11 ! − 10 ! 9 ! = 10 ! × ( 11 − 1 ) 9 ! = 10 × 10 = 100 \frac{11!-10!}{9!} = \frac{10!\times (11-1)}{9!}=10\times 10 = 100

Christopher Boo - 4 years, 4 months ago

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@Christopher Boo – How do you write in the 'box' like you've used above to show the L a T e X LaTeX code for the expression?

Tapas Mazumdar - 4 years, 3 months ago

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@Tapas Mazumdar – Wrap your expression with 3 back quote (`)

Christopher Boo - 4 years, 3 months ago

We can use subtraction of fractions to get 11 ! − 10 ! 9 ! \frac{11!-10!}{9!} = 11 ! 9 ! \frac{11!}{9!} - 10 ! 9 ! \frac{10!}{9!} = 110 - 10 = 100.

47 Helpful 0 Interesting 0 Brilliant 0 Confused

Could you explain how you calculated the values of 11 ! 9 ! \dfrac{11!}{9!} and 10 ! 9 ! \dfrac{10!}{9!} ?

Pranshu Gaba - 4 years, 3 months ago

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11! = 11 * 10 * 9!, cancel the denominator, etc.

Richard Desper - 4 years, 3 months ago

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It's not clear what are you typing.

Christopher Boo - 4 years, 3 months ago

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@Christopher Boo – He meant: 11! = 11×10×9! Analogously, 10! = 10×9! Then the canceling of denominators is ovious.

Martin Ramsch - 3 years, 4 months ago

Well, All roads lead to Rome.

Darmawan Putra Wijaya - 4 years, 4 months ago
Erfan Huq
Feb 10, 2017

take 9! as a factor from the numerator. then you get 9 ! ( 11 ∗ 10 − 10 ) 9 ! \frac{9!(11*10-10)}{9!} Then proceed to cancel the 9! out. then do the math

31 Helpful 0 Interesting 0 Brilliant 0 Confused

Right. We can write 11 ! 11! as 11 × 10 × 9 ! 11\times 10\times 9! and 10 ! 10! as 10 × 9 ! 10 \times 9! . We see that 9 ! 9! is common to both terms in the numerator, so we can factor it out.

Pranshu Gaba - 4 years, 3 months ago
Michael Fuller
Feb 22, 2017

11 ! − 10 ! 9 ! = 10 ! ( 11 − 1 ) 9 ! = 10 ! 9 ! ⋅ 10 = 10 ⋅ 10 = 100 . \dfrac{11!-10!}{9!} = \dfrac{10!(11-1)}{9!}= \dfrac{10!}{9!}\cdot 10 = 10 \cdot 10 = \large \color{#20A900}{\boxed{100}}.

10 Helpful 0 Interesting 0 Brilliant 1 Confused

Neat LaTeX!

The important observation here is to notice the relation n ! = n × ( n − 1 ) ! n! = n \times (n-1)! . Then everything becomes much more simple.

Christopher Boo - 4 years, 3 months ago
Ankur Shukla
Feb 23, 2017

As its given in the hints that,

n ! = n × ( n − 1 ) × ( n − 2 ) . . . . . 2 × 1 n! = n \times (n-1) \times (n-2) .....2 \times 1

So we can write,

10 ! = 10 × 9 × 8....2 × 1 10! = 10 \times 9 \times 8 .... 2 \times 1

i.e.

10 ! = 10 × 9 ! 10! = 10 \times 9!

By same factorisation we can get,

11 ! = 11 × 10 × 9 ! 11! = 11 \times 10 \times 9!

Putting this in the expression given in the question,

11 ! − 10 ! 9 ! \frac{11! - 10!}{9!} = 11 × 10 × 9 ! − 10 × 9 ! 9 ! \frac{11 \times 10 \times 9! - 10 \times 9!}{9!}

that implies

11 ! − 10 ! 9 ! \frac{11! - 10!}{9!} = 11 × 10 − 10 = 10 × 10 = 100 = 11 \times 10 - 10 = 10 \times 10 = 100

4 Helpful 0 Interesting 0 Brilliant 0 Confused

A lot of solutions skipped the explanation on how to simplify the expression. Thumbs up for your clear solution!

Christopher Boo - 4 years, 3 months ago
Prokash Shakkhar
Feb 16, 2017

\frac{11! - 10!}{9!} → \frac{11*10! - 10!}{9!} → \frac{10!*10}{9!} → \frac{10*9!*10}{9!} → \boxed{\color\red{100}}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Great! I see that you have used the recursive definition of factorials n ! = n × ( n − 1 ) ! n! = n \times (n-1)! twice to simplify the problem. 11 ! = 11 × 10 ! 11! = 11 \times 10! and 10 ! = 10 × 9 ! 10! = 10 \times 9!

Pranshu Gaba - 4 years, 3 months ago
Andra Sai Swaroop
Feb 27, 2017

take 10! common out of the whole equation you get 10!(11-1)/9! = 10! 10/9! = 10 10 = 100.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I believe this was #1 on AMC 10A 2016.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Yup, you're right.

The author should actual give credit to the source.

Pi Han Goh - 4 years, 3 months ago
Lance Fernando
Feb 22, 2017

(11! - 10!) / 9! = (11! / 9!) - (10! / 9!) = ((11)(10)) - 10 = 110 - 10 = 100

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you explain how you got (11! / 9!) to ((11)(10))?

Christopher Boo - 4 years, 3 months ago

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11! / 9! = (11x10 x 9 x...x 1) / (9 x 8 x 7 x...x 1). At this point, we can cancel out 9! and leave out 11 x 10.

Lance Fernando - 4 years, 3 months ago
Mohd Aasif
Feb 22, 2017

( 11! - 10! ) / 9!

= ( 11 x 10 x 9! -10 x 9! ) / 9!
= (9!) x (11 x 10 -10 ) / 9! 9! will be cancel from denominator as well as nominator
= 11 x 10 - 10
= 110 - 10
= 100



1 Helpful 0 Interesting 0 Brilliant 0 Confused

Hmm, there's seems to be some glitch in your presentation. Some digits are italicised, but some aren't. You might need to check again.

Christopher Boo - 4 years, 3 months ago
Ryan Howe
Feb 20, 2017

11!-10!/9! = 11x10x9!-10x9!/9!=11x10-10/1=110-10=100

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Careful there, you need to insert the appropriate brackets. In this case, you should start with (11! - 10!)/9, otherwise, we will interpret 11! - 10!/9! as 11 ! − 10 ! 9 ! 11! - \frac{10!}{9!} .

Pi Han Goh - 4 years, 3 months ago

Identity

(N+1)! - N! = N^2 × (N-1)!

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Right. To clarify, you should mention that it works for all positive integers N.

Otherwise, if we substitute N=0, we get an absurd equation.

Pi Han Goh - 4 years, 3 months ago
Akashdeep Roy
Jun 17, 2017

11!-10! can be written as 11X10!-10!. Take 10! common so it will become 10!(11-1)=10!X10. Now 10!X10 can be written as 10X9!X10. Since in the denominator 9! is present, 9! will cancel out and only 10X10 will remain which is equal to 100.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Factor out all of the numbers 9! and below that 11! and 10! have in common. (All of those numbers are common factors.) We're left with 11x10=110 for 11! and 10x1=10 for 10! 110-10 = 100. The answer is: 100. Remember: 9! was also canceled out in the denominator.

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