Question

I was doing a probability and statistics worksheet and I got stuck on a basic mathematical thing involving factorials. Could anyone explain why
x!(x+1) is the same as (x+1)! ?

Answer 1

sure

Answer 2

ok thanks

Answer 3

lets pick n = 6
then \(6!=6\times 5\times 4\times 3\times 2\times 1\) and \(n+1=7\) now \(7!=7\times 6\times 5\times 4\times 3\times 2\times 1=7\times 6!\)

Answer 4

yes I worked that out using wolfram. But the thing is, in an exam how would I realise this. Is there a specific method to simplifying factorials or is it just by intuition?

Answer 5

i hate to say that it is obvious, but it is obvious
consider any number n then \((n+1)!=(n+1)\times n\times (n-1)\times ...\times 2\times 1\) and this is fairly clearly the same as
\[(n+1)\times n!\]

Answer 6

sometimes factorials are even defined that way
\[1!=1, n!=n(n-1)!\]

Answer 7

you might also notice that \(\frac{(n+1)!}{n!}=n+1\) for example

Answer 8

I know these look obvious but I cant see it. lol

Answer 9

really try it with numbers and it will be clear

Answer 10

like the n(n-1)! = n! I dont see them as equal unless I start plugging in values using a calculator

Answer 11

how to you get 4! ? by multiplying 4*3*2*1
how do you get 5! ? by multiplying 5*4*3*2*1

Answer 12

so they are the same except for the 5

Answer 13

so basically there is no specific method of simplifying ?

Answer 14

don't think about what the final answer is, just think about what you are computing

Answer 15

\[\frac{7!}{5!}=\frac{7\times 6\times 5\times 4\times 3\times 2\times 1}{5\times 4\times 3\times 2\times 1}=7\times 6\]

Answer 16

with a little practice you would know right away that \(\frac{7!}{5!}=7\times 6\)

Answer 17

ok. Think i'm more confident with these now with your help. Just need more practice

Answer 18

Thanks for your help :)

Answer 19

yw