Question

Prove that.,
$\large \boxed{ \sf 0! = 1}$
​

Answer 1

★ Concept :-

Here the concept of Permutations has been used. We see that we have to prove if 0! = 1. If we have studied Permutations and Combinations then we already know some formulas. First of the formula is the standard formula of Permutations and the other two formulas are derivatives of the number patterns. So let's prove this question using those three formulas.

Let's do it !!

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★ Formula Used :-

$\;\boxed{\sf{\pink{^{n}P_{r}\;=\;\bf{\dfrac{n!}{(n\:-\:r)!}}}}}$

$\;\boxed{\sf{\pink{n!\;=\;\bf{\dfrac{(n\:+\:1)!}{(n\:+\:1)}}}}}$

$\;\boxed{\sf{\pink{n!\;=\;\bf{n\:\times\:(n\:-\:1)!}}}}$

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★ Solution :-

Let's try proving the given expression by each formula mentioned here seperately.

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~ Proof using first formula ::

We know that,

$\;\sf{\rightarrow\;\;^{n}P_{r}\;=\;\bf{\dfrac{n!}{(n\:-\:r)!}}}$

Let n = 1 and r = 1 here . So,

$\;\sf{\rightarrow\;\;^{n}P_{r}\;=\;\bf{\dfrac{1!}{(1\:-\:1)!}}}$

$\;\sf{\rightarrow\;\;^{n}P_{r}\;=\;\bf{\dfrac{1!}{0!}}}$

By defination of Permutations we know that r can never be greater than n . Here since r = n , so Permutation will also be 1 . So,

$\;\sf{\rightarrow\;\;1\;=\;\bf{\dfrac{1!}{0!}}}$

• Since 1! = 1

$\;\sf{\rightarrow\;\;1\;=\;\bf{\dfrac{1}{0!}}}$

$\;\sf{\rightarrow\;\;0!\;=\;\bf{\dfrac{1}{1}}}$

$\;\large{\boxed{\bf{\rightarrow\;\;\green{0!\;=\;\bf{1}}}}}$

Hence, proved.

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~ Proof using second formula ::

We know that,

$\;\sf{\rightarrow\;\;n!\;=\;\bf{\dfrac{(n\:+\:1)!}{(n\:+\:1)}}}$

• Let n = 0

$\;\sf{\rightarrow\;\;0!\;=\;\bf{\dfrac{(0\:+\:1)!}{(0\:+\:1)}}}$

$\;\sf{\rightarrow\;\;0!\;=\;\bf{\dfrac{(1)!}{(1)}}}$

• Since 1! = 1

$\;\sf{\rightarrow\;\;0!\;=\;\bf{\dfrac{1}{1}}}$

$\;\large{\boxed{\bf{\rightarrow\;\;\red{0!\;=\;\bf{1}}}}}$

Hence, proved .

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~ Proof using third formula ::

We know that,

$\;\sf{\rightarrow\;\;n!\;=\;\bf{n\:\times\:(n\:-\:1)!}}$

• Let n = 1

$\;\sf{\rightarrow\;\;1!\;=\;\bf{1\:\times\:(1\:-\:1)!}}$

$\;\sf{\rightarrow\;\;1!\;=\;\bf{1\:\times\:0!}}$

$\;\sf{\rightarrow\;\;0!\;=\;\bf{\dfrac{1!}{1}}}$

• Since 1! = 1

$\;\sf{\rightarrow\;\;0!\;=\;\bf{\dfrac{1}{1}}}$

$\;\large{\boxed{\bf{\rightarrow\;\;\blue{0!\;=\;\bf{1}}}}}$

Hence, proved.

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★ More to know :-

$\;\tt{\leadsto\;\;^{n}C_{r}\;=\;\dfrac{n!}{(n\:-\:r)!}}$

Answer 2

Answer:

Originally answered: how can I prove 0!=1

The factorial of a number is defined as:

= n! = n*(n-1)*(n-2)*(n-3)*...*1n !

= n*(n-1)*(n-2)*(n-3)*...*1

So,

5!=5*4*3*2*1 = 1250! = 5*4*3*2*1 = 120

4! = 4*3*2*1 = 244! = 4*3*2*1 = 24

Hence, it can be observed that:

n! = n*(n-1)! n! = n*(n-1)!

From this equation, substituting 1 from:

1! = 1*(1-1)! 1! 1! = 1

1! = 1*0! 1! = 1*0!

0! = 1/10! = 1/1

0! = 1

It's proved

Step-by-step explanation:

hope it will help you