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@Mahdi Raza
–
I was about to comment the proof-related to combinations after I saw Percy's proof, but then I saw you writing this (sad noises) LOL. I will write the proof-related to permutations then.

I assume the reader knows, about permutations and a little about factorials.

The formula for no. of permutations is ${ _{ n }{ P }_{ k } }=P(n,k)=\frac { n! }{ (n-k)! }$ or basically if we take any $k$ objects from $n$ objects at a time, then how many ways can the $k$ objects be arranged is the number of permutations.

So, if we take $n$ objects from $n$ objects and ask for the number of ways we can arrange them, basically we are asking how many ways can all the $n$ objects be arranged which by a primary definition of factorials is n! So using permutations we can write as,

A common-sensical(Yeah, I just invented that word, deal with it) approach would be that 0! is 0 multiplied 0 times because 1 is greater than 0. 0 multiplied 0 times is $0^{0}$, aka 1 :) @SRIJAN Singh@Siddharth Chakravarty

What do you even mean? I would suggest the common-sensical way, Lol that 0! Means how many ways can we arrange 0 objects which should be 1 i.e there is no way or to do nothing.

$n! = n(n-1)(n-2) \ldots 1$ or $n!$ is product of all numbers less than n and greater than 1. There are 0 ways to reach 1 from 0 like that $0! = 0^{0} = 1$. THis is in a vague sense @Siddharth Chakravarty

@Percy Jackson
–
Did you just defy yourself because you said product of all number less than n and greater than 1, and 0 has no number less than it which is greater than 1? Basically the factorial notation was brought into due to permutations and combinations as I said in the comment.

What's with the notifications? I got like 10 of them in this note, leading nowhere, are you guys sending and deleting comments @Siddharth Chakravarty and @SRIJAN Singh ????

@Percy Jackson
–
yah i also if i hover to notification button and then click ,if i get bore scrolling comments than CLICK DISMISS ALL LOL USED IF many times

$n!$ is the number of ways you can order a set with $n$ elements, and the only way to order an empty set (with $0$ elements) is the empty set {} itself.

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## Comments

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TopNewest@Mahdi Raza

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0! = 1 :

1! = 1

2! = 1!*2

2! = 2

3! = 2!*3

3! = 6

4! = 3!*4

4! = 24

turn this around:

4! = 24

3! = 4!/4

3! = 6

2! = 3!/3

2! = 2

1! = 2!/2

1! = 1

0! = 1!/1

0! = 1

@SRIJAN Singh

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Yeah, same reason!

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Its the only proof I know for 0! @Mahdi Raza, do you know any others?

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$n \choose 0$. The ways of choosing 0 objects from a group of $n$ objects is $\boxed{1}$. Hence:

There was one related to$\begin{aligned} n \choose 0 &= \dfrac{n!}{(n-0)!0!} \\ \\ 1 &= \dfrac{n!}{n!0!} \\ \\ 0! &= \dfrac{\cancel{n!}}{\cancel{n!}} \\ \\ 0! &= \boxed{1} \end{aligned}$

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@Mahdi Raza, Thanks :)

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@Percy Jackson nice

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Thanks :)

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I assume the reader knows, about permutations and a little about factorials.

The formula for no. of permutations is ${ _{ n }{ P }_{ k } }=P(n,k)=\frac { n! }{ (n-k)! }$ or basically if we take any $k$ objects from $n$ objects at a time, then how many ways can the $k$ objects be arranged is the number of permutations.

So, if we take $n$ objects from $n$ objects and ask for the number of ways we can arrange them, basically we are asking how many ways can all the $n$ objects be arranged which by a primary definition of factorials is n! So using permutations we can write as,

$n!=\frac { n! }{ (n-n)! }$

Thus, $n!=\frac { n! }{ 0! }$

By simplifying, we get 0!=1.

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A common-sensical(Yeah, I just invented that word, deal with it) approach would be that 0! is 0 multiplied 0 times because 1 is greater than 0. 0 multiplied 0 times is $0^{0}$, aka 1 :) @SRIJAN Singh @Siddharth Chakravarty

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What do you even mean? I would suggest the common-sensical way, Lol that 0! Means how many ways can we arrange 0 objects which should be 1 i.e there is no way or to do nothing.

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$n! = n(n-1)(n-2) \ldots 1$ or $n!$ is product of all numbers less than n and greater than 1. There are 0 ways to reach 1 from 0 like that $0! = 0^{0} = 1$. THis is in a vague sense @Siddharth Chakravarty

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For a (possibly) more intuitive explanation, recall that $n!$ may be interpreted as the number of ways to arrange $n$ distinct objects in a line.

For example, $2!=2$ and we can arrange $\square$ and $\blacksquare$ in 2 ways: $\square\blacksquare$ and $\blacksquare\square$.

$1!=1$, and there is one way to arrange the single item $\blacksquare$, like this: $\blacksquare$.

So $0!$ is the number of ways to arrange nothing. There is one way to arrange nothing in a line, like this:

So $0!=1$, though I admit that the interpretation of arranging "nothing" is a little philosophical.

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What's with the notifications? I got like 10 of them in this note, leading nowhere, are you guys sending and deleting comments @Siddharth Chakravarty and @SRIJAN Singh ????

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i got 34 comments a day

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did it ever go to 100+? I left brilliant for one day, I had 100+ notifications :)

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@Yajat Shamji.do this

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@Mahdi Raza. do this

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@Siddharth Chakravarty @SRIJAN Singh -

## -_- Why are you making so many notifications and deleting your comments?!?!?!?!?!??!?!

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To trouble you. LOL :) Actually, I am not deleting @SRIJAN Singh did, he had the main comment.

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$\LARGE \textsf{Unsubscribe :)}$

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$\text{\large I think to do the same, but I am not doing, incase somebody comments something useful, notifications don't worry much}$

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$\Large \textsf{Why am I still getting notifications?}$

$\Large \textsf{I already Unsubbed this note.........}$

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@Percy Jackson why are you making drawings of fullstop lol

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@Siddharth Chakravarty,@Percy Jackson because siddharth had given the answers that was asked by me

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Why delete comments then????????????????????????????????????????????

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i asked some of the questions like how was your naest exam and ask him to do my latest problems

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@Percy Jackson,@Siddharth Chakravarty.do this

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Note: $\displaystyle n! = \Gamma(n+1) = \int_{0}^{\infty}t^n e^{-t}dt$

$\begin{aligned} 0! = \Gamma(1) = \int_{0}^{\infty}t^0 e^{-t}dt &= \int_{0}^{\infty}e^{-t}dt \\ &= -e^{-t}\bigg|^{\infty}_{0} \\ &= -e^{-\infty} -(-e^0) = 0 + 1 = \boxed{1} \end{aligned}$

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Thanks for explanation.

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$n!$ is the number of ways you can order a set with $n$ elements, and the only way to order an empty set (with $0$ elements) is the empty set {} itself.

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...hi @SRIJAN Singh, you actually look pissed

What is going on with you both

(referring to NEED TO NOTIFY THE STAFF? COMMENT BELOW)If you feel this comment is offensive, I will delete this comment, k?

I just wanna know what's happening between you and Andrew

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nothin much he's tryin to become big daddy and treats me like i m his target idk why?

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