Question

prove that 0!=1
please explain also.​

Answer 1

Answer:

0!=1

Step-by-step explanation:

TO PROVE

n!=n(n-1)

n=n!/(n-1)

putting n=1/ ,so,

1=1/0!

0!=1/1

∴0!=1

hope this helps u......

Answer 2

Answer:

There are many was to prove this :

1> Let us consider some examples :

5! = 5*4*3*2*1 = {6*5*4*3*2*1} ÷ 6 = 6! ÷ 6

4! = 4*3*2*1 = {5*4*3*2*1} ÷ 5 = 5! ÷ 5

3! = 3*2*1 = {4*3*2*1} ÷ 4 = 4! ÷ 4

DID YOU NOTICE THE PATTERN?

n! = (n+1)! ÷ (n+1) GOT IT ?

Following this pattern,

when we put n=0, we get :

0! = (0+1)! ÷ (0+1) = 1! ÷ 1 = 1 ÷ 1 = 1

→ 0! = 1 (proved)

2> Let us consider some examples :

4! = 4*3*2*1 = 4 * 3!

3! = 3*2*1 = 3 * 2!

2! = 2*1 = 2 * 1!

DID YOU NOTICE THE PATTERN?

n! =n * (n-1)! GOT IT ?

putting n = 1, we get

1! = 1 * (1–1)! = 1 * 0!

→ 0! = (1! / 1) = (1 / 1) = 1

→ 0! = 1 (proved)

The above methods were INDIRECT PROOFS (PROVED JUST BY OBSERVING THE NUMBER PATTERNS )

3> to 5> LOGICAL PROOFS (RECOMMENDED)

3> PERMUTATION :

[math] ^nP_r [/math]= No. of ways of ARRANGING ‘n’ objects in ‘r’ places =[math] [/math]n! / (n-r)!

if we put n=1 and r = 1 then RHS = 1! / (1–1)!= 1 / 0!

LHS = [math] ^1P_1[/math] = 1 BECAUSE 1 OBJECT CAN BE PLACED(ARRANGED) IN 1 SEAT IN 1 WAY ONLY

Thus 1 = 1 / 0! → 0! = 1/1 → 0! = 1 (proved)

4> SET LOGIC:

You can justify the above argument by considering arrangemnent of elements in a set (NOTE : During “ARRANGEMENT”(PERMUTATION) OF ELEMENTS, ORDER MATTERS BUT DURING “SELECTION”(COMBINATION) OF ELEMENTS, ORDER DOES NOT MATTER) :-

a set having two elements viz. ‘x’ and ‘y,’ can be arranged in 2 (i.e. 2!) ways :

{x,y} or {y,x}

a set having three elements viz. ‘x’, ‘y’ and ‘z,’ can be arranged in 6 (i.e. 3!)ways :

{x,y,z} or {x,z,y} or {y,x,z} or {y,z,x} or {z,x,y} or {z,y,x}

Analogically, a set having zero elements can be arranged in 1(i.e. 0!) way: {} (null set / void set / empty set)

BECAUSE ZERO ELEMENTS CAN BE SHOWN BY ONY A NULL SET WHICH IS ALSO A SET (AN ARRANGEMENT) AND HENCE WE SHOULD COUNT IT TOO!

So, only one set i.e. one arrangement

5>COMBINATION:

[math] ^n[/math]C[math]_r [/math]= No. of ways of SELECTING ‘r’ objects from ’n’ objects =[math] [/math]n! / r!(n-r)!

if we put n=1 and r = 1 then RHS = 1! /1! (1–1)! = 1 /(1 * 0!) = 1 / 0!

LHS =[math] ^1[/math]C[math]_1[/math] = 1 BECAUSE 1 OBJECT CAN BE SELECTED FROM 1 OBJECT IN 1 WAY ONLY

Thus 1 = 1 / 0! → 0! = 1/1 → 0! = 1 (proved)

///////////////Hope it helps///////////

/////////Please mark me brainliest//////////////