Question

Could someone show me how to simplify this.
(n-4)!/(n-5)!

Answer 1

@AkashdeepDeb @madrockz

Answer 2

Good question!

\[\frac{ (n-4)! }{ (n-5)! } \]

= \[\frac{ (n-4)(n-5)(n-6)(n-7).........3.2.1 }{ (n-5)(n-6)(n-7).........3.2.1 }\]

This is because factorials always reduce by 1!
NOTE: n! = n(n-1)(n-2)(n-3)(n-4)...........3.2.1

So now cancel (n-5)!

= \[\frac{ (n-4)(n-5)(n-6)(n-7).........3.2.1 }{~~~~~~~~~~~~~ (n-5)(n-6)(n-7).........3.2.1 }\]
= \[\frac{(n-4)(n-5)!}{~~~~~~~~~~~~(n-5)!}\]

It leaves you with

\[(n-4)\]

Understood? :)

Answer 3

Okay, I would like help with one more in that step by step format you helped me in.... (n+4)!/(n+2)! and the problem (n+2)!/n!

Answer 4

@AkashdeepDeb

Answer 5

If you understood this step by step format.
Try the others yourself.
I'll help.

Answer 6

ok

Answer 7

for (n+4)!/(n+2)! I got (n+2)(n+3)

Answer 8

@AkashdeepDeb

Answer 9

Did I do it correctly? @AkashdeepDeb

Answer 10

No see
(n+4)! = (n+4)(n+3)(n+2)(n+1)n..............3.2.1
(n+2)! = (n+2)(n+1)n......3.2.1

Now divide them.
What do you get?

Answer 11

(n+2)?

Answer 12

n-4

Answer 13

No.

\[\frac{ (n+4)(n+3)(n+2)(n+1)n..............3.2.1 }{ (n+2)(n+1)n......3.2.1 }\]

DIVIDE!

\[\frac{ (n+4)(n+3)(n+2)(n+1)n..............3.2.1 }{ ~~~~~~~~~~~~~~~~~~~~~(n+2)(n+1)n................3.2.1 }\]

(n+2)! cancels out.

Leaving us with
\[(n+4)(n+3)\]

Understood? :)
Get the next one right! :COME ON! :D

Answer 14

@BaileyC
Did you **UNDERSTAND** this or not?

Answer 15

Yes, i think i am solving the next question. please let me know when I am done solving it how i did

Answer 16

i just told you

Answer 17

Okay.

Answer 18

(n+2)!/n! = (n+2)(n+1)

Answer 19

Good! :)

Answer 20

I really appreciate the help. R u a math teacher?

Answer 21

Well! I am both. I learn while I teach. :)
But not like a Pro Math teacher who earns! :P

Answer 22

Whoa really? cool.

Answer 23

Okay I have figured out the simplifying of expressions... Now I have a question on solving expressions.


So I have the question here,
n/(n-2)!=72

How do I solve for n with another number variable?
@AkashdeepDeb

Answer 24

Solve the LHS like you solved simply.
And tell me the LHS.

Answer 25

The answer is just 9 but I don't remember any of the steps to come to the single number! @AkashdeepDeb

Answer 26

DO NOT MUG UP STEPS!

Try to solve it by understanding factorials.

Answer 27

hmm what do you mean

Answer 28

http://www.mathsisfun.com/numbers/factorial.html

Answer 29

I just dont get it algebraically

Answer 30

n!/(n-2)!

n(n-1)(n-2)!/(n-2)! [THIS IS THE FOURTH TIME]
Plz SEE THE WEBSITE, and get your basics cleared.

n(n-1) = 72
n(n-1) = 9*8

Thus n=9

Answer 31

your mean

Answer 32

*you're

I speak the truth.

See if you don't understand what I say, why don't you just ask me then?
This is the 4th similar factorial problem.
I am ready to help.

Go through the website and I am sure you'll be able to solve these. :)

Answer 33

I hope your video loses

Answer 34

@AkashdeepDeb

Answer 35

OMG!

Your curses aren't helping you.
Neither are they moving me. :P
Do you want to understand? :D

Answer 36

bye