Why is 0!=1 0! = 1 ?

Note by Srijan Singh
2 months, 3 weeks ago

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@Mahdi Raza

Yajat Shamji - 2 months, 3 weeks ago

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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

Percy Jackson - 2 months, 3 weeks ago

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Yeah, same reason!

Mahdi Raza - 2 months, 3 weeks ago

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

Percy Jackson - 2 months, 3 weeks ago

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@Percy Jackson There was one related to (n0)n \choose 0. The ways of choosing 0 objects from a group of nn objects is 1\boxed{1}. Hence:

(n0)=n!(n0)!0!1=n!n!0!0!=n!n!0!=1\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}

Mahdi Raza - 2 months, 3 weeks ago

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@Mahdi Raza Cool @Mahdi Raza, Thanks :)

Percy Jackson - 2 months, 3 weeks ago

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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.

Siddharth Chakravarty - 2 months, 3 weeks ago

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@Percy Jackson nice

SRIJAN Singh - 2 months, 3 weeks ago

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Thanks :)

Percy Jackson - 2 months, 3 weeks ago

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

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

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

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

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

By simplifying, we get 0!=1.

Siddharth Chakravarty - 2 months, 3 weeks ago

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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 000^{0}, aka 1 :) @SRIJAN Singh @Siddharth Chakravarty

Percy Jackson - 2 months, 3 weeks ago

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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.

Siddharth Chakravarty - 2 months, 3 weeks ago

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n!=n(n1)(n2)1n! = n(n-1)(n-2) \ldots 1 or n!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!=00=10! = 0^{0} = 1. THis is in a vague sense @Siddharth Chakravarty

Percy Jackson - 2 months, 3 weeks ago

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@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.

Siddharth Chakravarty - 2 months, 2 weeks ago

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

For example, 2!=22!=2 and we can arrange \square and \blacksquare in 2 ways: \square\blacksquare and \blacksquare\square.

1!=11!=1, and there is one way to arrange the single item \blacksquare, like this: \blacksquare.

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

So 0!=10!=1, though I admit that the interpretation of arranging "nothing" is a little philosophical.

Matthew Christopher - 2 months, 2 weeks ago

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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 ????

Percy Jackson - 2 months, 2 weeks ago

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i got 34 comments a day

SRIJAN Singh - 2 months, 2 weeks ago

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

Percy Jackson - 2 months, 2 weeks ago

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@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

SRIJAN Singh - 2 months, 2 weeks ago

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@Yajat Shamji.do this

SRIJAN Singh - 2 months, 2 weeks ago

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@Mahdi Raza. do this

SRIJAN Singh - 2 months, 2 weeks ago

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@Siddharth Chakravarty @SRIJAN Singh -

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

Percy Jackson - 2 months, 2 weeks ago

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

Siddharth Chakravarty - 2 months, 2 weeks ago

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Unsubscribe :)\LARGE \textsf{Unsubscribe :)}

Percy Jackson - 2 months, 2 weeks ago

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

Siddharth Chakravarty - 2 months, 2 weeks ago

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@Siddharth Chakravarty Why am I still getting notifications?\Large \textsf{Why am I still getting notifications?}

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

Percy Jackson - 2 months, 2 weeks ago

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@Percy Jackson because this note loves you

SRIJAN Singh - 2 months, 2 weeks ago

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@Srijan Singh

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Percy Jackson - 2 months, 2 weeks ago

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@Percy Jackson @Percy Jackson why are you making drawings of fullstop lol

SRIJAN Singh - 2 months, 2 weeks ago

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

SRIJAN Singh - 2 months, 2 weeks ago

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

Percy Jackson - 2 months, 2 weeks ago

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

SRIJAN Singh - 2 months, 2 weeks ago

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@Percy Jackson,@Siddharth Chakravarty.do this

SRIJAN Singh - 2 months, 2 weeks ago

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

0!=Γ(1)=0t0etdt=0etdt=et0=e(e0)=0+1=1 \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}

James Watson - 4 weeks ago

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Thanks for explanation.

SRIJAN Singh - 3 weeks, 6 days ago

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n!n! is the number of ways you can order a set with nn elements, and the only way to order an empty set (with 00 elements) is the empty set {} itself.

Lâm Lê - 3 weeks, 4 days ago

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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 (yes, that is his name...)

Frisk Dreemurr - 1 week ago

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

SRIJAN Singh - 1 week ago

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