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Q: Why factorial of 0 equals 1?

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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

Factorial(0), or 0! = 1.

The trick is that zero factorial (0!) equals 1[1], so (0!+0!+0!+0!+0!)! = 5! = 120

145 1! = 1 4! = 24 5! = 120

25 factorial equals 15,511,210,043,330,985,984,000,000

yes, 0!=1 default.

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

The simple answer is that it is defined to be 1. But there is reason behind the decision.As you know, the factorial of a number (n) is equal to:n! = n * (n-1) * (n-2) ... * 1Another way of writing this is:n! = n * (n-1)!Suppose n=1:1! = 1 * 0!or1 = 1 * 0!or1 = 0!So by defining 0! as 1, formula involving factorials will work for all integers, including 0.

simply, any number divided by 0 is 0.

A recursive formula for the factorial is n! = n(n - 1)!. Rearranging gives (n - 1)! = n!/n, Substituting 'n - 1' as 0 -- i.e. n = 1 -- then 0! = 1!/1, which is 1/1 = 1.

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

3! = 3×2×1 = 6

1 factorial = 1

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

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.

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.

== == 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--; } } }

factorial of -1

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

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

A C, C#/Java code segment:int Factorial(int n) {if (n

AnswerAnswer: ( 0! + 0! + 0! + 0! + 0! ) ! = 120 Explanation: Here we have used operator called " factorial ". As you know that 0! = 1 so, = ( 0! + 0! + 0! + 0! + 0! ) ! = ( 1 + 1 + 1 + 1 + 1 ) ! = (5 )! = 120 : ( 0! + 0! + 0! + 0! + 0! ) ! = 120 Explanation: Here we have used operator called " factorial ". As you know that 0! = 1 so, = ( 0! + 0! + 0! + 0! + 0! ) ! = ( 1 + 1 + 1 + 1 + 1 ) ! = (5 )! = 120

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

Zero factorial = 1