Question

How do I solve N!/(N-1)!?

Answer 1

Do you understand what I'm doing in this example here?
\[\Large \frac{5!}{4!}\quad=\quad \frac{5\cdot4\cdot3\cdot2\cdot1}{4!}\quad=\quad \frac{5\cdot4!}{4!}\quad=\quad 5\]

Answer 2

yep

Answer 3

\[\Large (5-1)!\quad=\quad 4!\]

So we could rewrite our example as:\[\Large \frac{5!}{4!}\quad=\quad \frac{5!}{(5-1)!}\]

Answer 4

hmm...

Answer 5

an interesting development...

Answer 6

lol :)

Answer 7

so

Answer 8

ehh

Answer 9

I now the 1x2x3... in both numerators and denominators are canceled out

Answer 10

I have this

Answer 11

\[1\times 2 \times 3...(n)/1 \times 2 \times 3...(n-1)\]

Answer 12

So when n=3,\[\Large \frac{n}{(n-1)!}\quad=\quad \frac{3\cdot2\cdot1}{2\cdot1}\quad=\quad 3\]
When n=4,\[\Large \frac{n}{(n-1)!}\quad=\quad \frac{4\cdot3\cdot2\cdot1}{3\cdot2\cdot1}\quad=\quad 4\]What number are we ending up with in each case?

Answer 13

woops those numerators should be n!, sorry.

Answer 14

n

Answer 15

Yesss good good.

Answer 16

so how would I write an answer out in variables?

Answer 17

(n)(n-1)/(n-1)?

Answer 18

Here is another way to think about it.
If we take the first number out of that factorial in the numerator (similar to what we did n the example, we're left with this:
\[\Large \frac{5\cdot(5-1)!}{(5-1)!}\]

Answer 19

Ah sorry lemme throw the n back in.

Answer 20

\[\Large \frac{n!}{(n-1)!}\quad=\quad \frac{n\cdot(n-1)!}{(n-1)!}\]

Answer 21

Then we have a nice cancellation from there, yes? :o

Answer 22

but how do we get the (n-1) ontop?

Answer 23

(n-1) in the numerator that is?

Answer 24

like, for a problem like (n-1)!/(n-3)!, would the answer be (n-1)(n-3)/(n-3)?

Answer 25

sorry for all the questions, I hope I am not bugging you

Answer 26

\[\Large \frac{(n-1)!}{(n-3)!}\]In this example, let's try to play around with the numerator because it starts from a larger number.

Look at this example a sec:\[\Large 4!\]If I wanted to pull the 4 out of the factorial, the what does it leave us with in factorial form? The next number below 4, yes?\[\Large 4!\quad=\quad 4\cdot 3!\]Does that setup make sense? :o
I just pulled a 4 outside of the factorial thing.

Answer 27

yes

Answer 28

n=4 for example, would be (4-1)!=3!=3 times 2 times 1=6 in the numberator

Answer 29

(4-3)!=1!=1 in the denominator?

Answer 30

yes

Answer 31

6/1=6

Answer 32

So what we want to do is, we want to peel numbers off of the larger factorial, until it matches the smaller one.

Answer 33

hmm

Answer 34

so the answer wouldn't be 6n?

Answer 35

Mmm no not quite :) Stay with me a sec!

Answer 36

\[\Large \frac{(n-1)!}{(n-3)!}\quad=\quad \frac{(n-1)\cdot(?)!}{(n-3)!}\]So if we pulled the (n-1) out of the top factorial, what number should our factorial start from then?

It will be the next number below n-1, which is? :D

Answer 37

(n-3) :D!!!!!!!!!!!!!!

Answer 38

No no not quite. that is the number `2 below` n-1.
You jumped too far.

To get to the number `one below` n-1 we simply subtract 1 ok?
So it gives us n-2

Answer 39

\[\Large \frac{(n-1)!}{(n-3)!}\quad=\quad \frac{(n-1)\cdot(n-2)!}{(n-3)!}\]

Answer 40

:(, but at least I tried :D

Answer 41

XD

Answer 42

So now we want to deal with our new factorial on top, it's still too large.
We'll peel another number out of it.\[\Large \frac{(n-1)\cdot(n-2)!}{(n-3)!}\quad=\quad \frac{(n-1)\cdot(n-2)\cdot(?)!}{(n-3)!}\]So we pulled the largest number (n-2) out of the factorial.
What will our new factorial start from?
It would be the number just below n-2, which is? :O

Answer 43

(n-3) :DDD!!!!!!

Answer 44

Good good, that's (n-2)-1, just one number below n-2!

Answer 45

I think I got the answer

Answer 46

\[\Large =\quad \frac{(n-1)\cdot(n-2)\cdot\color{#CC0033}{(n-3)!}}{\color{#CC0033}{(n-3)!}}\]

Answer 47

n squared -3n+2

Answer 48

:D?

Answer 49

Yesss good job! :)

Answer 50

thank you so much

Answer 51

These problems can be a little tricky.
You start with two factorials and you need to try and match them up by shrinking down the larger one.

Answer 52

\c:/ np