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Q: What is the factorial of 0?

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Factorial(0), or 0! = 1.

Zero factorial, written as 0!, equals 1. This is a simple math equation.

0!=1! 1=1 The factorial of 0 is 1, not 0

(0!+0!+0!+0!+0!)!=120 !=factorial

simply, any number divided by 0 is 0.

Definition of FactorialLet n be a positive integer. n factorial, written n!, is defined by n! = 1 * 2 * 3 * ... (n - 1) * nThe special case when n = 0, 0 factorial is given by: 0! = 1

/* gcc -ansi -Wall -Wextra -pedantic -s -static 0.c -o 0 */ #include <stdio.h> int main ( ) { int n , factorial = 1 ; printf ( "enter the value of n\n") ; scanf ( "%i" , & n ) ; while ( n != 0 ) { factorial *= n ; n -- ; } printf ( "The factorial of n is\n%i\n" , factorial ) ; return 0; }

yes. (0!+0!+0!+0!+0!)! where ! refers the factorial of the number

yes. (0!+0!+0!+0!+0!)! where ! refers the factorial of the number

Zero factorial is one because n! = n-1! X n. For example: 4! = (4-1) X 4. If zero factorial was zero, that would mean 1! =(1-1) X 1 = 0 X 1=0. Then if 1!=0, then even 999! would equal zero. Therefore, zero factorial equals 1.

yes, 0!=1 default.

/* gcc -ansi -Wall -Wextra -pedantic -s -static 0.c -o 0 */ #include <stdio.h> int main ( ) { int n = 5 , factorial = 1 ; while ( n != 0 ) { factorial *= n ; n -- ; } printf ( "%i\n" , factorial ) ; return 0; }

Factorial (n) = n * Factorial (n-1) for all positive values n given Factorial (1) = Factorial (0) = 1. Pseudo-code: Function: factorial, f Argument: positive number, n IF n<=1 THEN RETURN 1 ELSE RETURN n * f(n-1) END IF

double factorial(double N){double total = 1;while (N > 1){total *= N;N--;}return total; // We are returning the value in variable title total//return factorial;}int main(){double myNumber = 0;cout > myNumber;cout

That is related with the fact that 1 is the identity element (or neutral element) of multiplication - and factorials are defined as multiplications. Defining 0 factorial thus simplifies several formulae.

/*this is the program to find the factorial to n numbers*/ #include<stdio.h> void main() { int n,f,factorial(int); printf("which number facrorial you want?"); scanf("%d",&n); f=factorial(n); printf("The factorial upto %d is %d",n,f); } int factorial(int a) { if(a==0) return 1; else return (a)*factorial(a-1); }

== == using recursions: unsigned int Factorial( unsigned int x) { if(x>0) { return ( x * Factorial(x-1)); } else { return(1); } } factorial: unsigned int Factorial( unsigned int x) { unsigned int u32fact = 1; if( x == 0) { return(1); } else { while(x>0) { u32fact = u32fact *x; x--; } } }

double Factorial(int i) { double x = 1; if (i <= 1) return 1; do { x *= i--; } while (i > 0); return x; }

a factorial number is a number multiplied by all the positive integers i.e. 4!=1x2x3x4=24 pi!=0.14x1.14x2.14x3.14 0!=1

What is the rationale for defining 0 factorial to be 1?AnswerThe defining 0 factorial to be 1 is not a rationale."Why is zero factorial equal to one?" is a problem that one has to prove.When 0 factorial to be 1 to be proved,the defining 0 factorial to be 1 is unvaluable.One has only one general primitive definition of a factorial number:n! = n x (n-1) x (n-2) x (n-3) x ... x 2 x 1.After that zero factorial denoted 0! is a problem that one has to acceptby convention 0!=1 as a part of definition.One has to prove zero factorial to be one.Only from the definition of a factorial number and by dividing both sidesby n one has: n!/n (n-1)! or (n-1)! = n!/nwhen n=2 one has (2-1)! = 2!/2 or 1! = 2x1/2 or 1! = 1when n=1 one has (1-1)! = 1!/1 or 0! = 1/1 or 0! = 1. =This is a proof that zero factorial is equal to one to be known.But a new proof is:A Schema Proof Without WordsThat Zero Factorial Is Equal To One.... ... ...Now the expression 0! = 1 is already a proof, not need a definitionnor a convention. So the defining 0 factorial to be 1 is unvaluable.The proof "without words" abovethat zero factorial is equal to one is a New that:*One has not to accept by convention 0!=1 anymore.*Zero factorial is not an empty product.*This Schema leads to a Law of Factorial.Note that the above schema is true but should not be used in a formal proof for 0!=1.The problem arises when you simplify the pattern formed by this schema into a MacLauren Series, which is the mathematical basis for it in the first place. Upon doing so you arrive with, . This representation illustrates that upon solving it you use 0!.In proofs you cannot define something by using that which you are defining in the definition. (ie) 0! can't be used when solving a problem within a proof of 0!.For clarification, the above series will represent the drawn out solution for the factorial of a number, i. (ie) 1×76 -6×66 +15×56 -20×46 +15×36 -6×26 +1×16 , where i=6.

Zero factorial is equal to one. 0! = 1

int main() { // Variable declarations. unsigned long int factorial = 1 , number = 1; // reads a number for finding its factorial. cout > number; while( number > 1 ) { factorial *= number * ( number - 1 ); number -= 2; } cout

The value of 9 factorial plus 6 factorial is 363,600

For any positive integer, n, factorial (n) can be calculated as follows: - if n<2, return 1. - otherwise, return n * factorial (n-1). The algorithm is recursive, where n<2 represents the end-point. Thus for factorial (5) we find the following recursive steps: factorial (5) = 5 * factorial (4) factorial (4) = 4 * factorial (3) factorial (3) = 3 * factorial (2) factorial (2) = 2 * factorial (1) factorial (1) = 1 We've now reached the end-point (1 is less than 2) and the results can now filter back up through the recursions: factorial (2) = 2 * factorial (1) = 2 * 1 = 2 factorial (3) = 3 * factorial (2) = 3 * 2 = 6 factorial (4) = 4 * factorial (3) = 4 * 6 = 24 factorial (5) = 5 * factorial (4) = 5 * 24 = 120 Thus factorial (5) = 120. We can also use a non-recursive algorithm. The factorial of both 0 and 1 is 1 thus we know that the return value will always be at least 1. As such, we can initialise the return value with 1. Then we begin iterations; while 1<n, multiply the return value by n and then subtract 1 from n. We can better represent this algorithm using pseudocode: Function: factorial (n), where n is an integer such that 0<=n. Returns an integer, f. Let f = 1 Repeat while 1<n Let f = f * n Let n = n - 1 End repeat Return f

#include #include using std::cin;using std::cout;using std::endl;using std::tolower;long factorial(int N);int main(){int N = 0; //factorial of Nchar command = 'n';do{cout > N;cout