prove that 0!=1 ( need solution of this question )
Answers
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 :
nPr = No. of ways of ARRANGING ‘n’ objects in ‘r’ places = n! / (n-r)!
if we put n=1 and r = 1 then RHS = 1! / (1–1)!= 1 / 0!
LHS = 1P1 = 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:
n C r = No. of ways of SELECTING ‘r’ objects from ’n’ objects = n! / r!(n-r)!
if we put n=1 and r = 1 then RHS = 1! /1! (1–1)! = 1 /(1 * 0!) = 1 / 0!
LHS = 1 C 1 = 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 helped :)