It is n! or n factorial.
P(n,r)=(n!)/(r!(n-r)!)This would give you the number of possible permutations.n factorial over r factorial times n minus r factorial
Factorial for number N is N x N-1 x N-2 X N- (N-1). e.g. if you need to calculate factorial for 5 then compute 5 x 4 x 3 x 2 x 1.
#include #include using std::cin;using std::cout;using std::endl;using std::tolower;long factorial(const int& N);int main(){int N = 0; //factorial of Nchar command = 'n';do{cout > N;cout
Since n! is the product of all the numbers from 1 through n and (n+1)! is everything in n! multiplied by n+1, the quotient is n+1 ■
It is not except when n = 1.
what is the value of negative n factorial ?
this is a code for calculating it recursivelly: float Factorial (float n) { if (n<=1) return 1.0; else return n* Factorial(n-1); }
' Iterative solution Function iterativeFactorial(ByVal n As Long) As Long Dim factorial As Long = 1 For i As Long = 1 To n factorial *= i Next Return factorial End Function ' Recursive solution Function recursiveFactorial(ByVal n As Long) As Long If n <= 1 Then Return n End If Return n * recursiveFactorial(n - 1) End Function
If you have N things and want to find the number of combinations of R things at a time then the formula is [(Factorial N)] / [(Factorial R) x (Factorial {N-R})]
// Iterative solution public static final long iterativeFactorial(final long n) { long factorial = 1; for (long i = 1; i <= n; i++) { factorial *= i; } return factorial; } // Recursive solution public static final long recursiveFactorial(final long n) { if (n <= 1) { return n; } return n * recursiveFactorial(n - 1); } // Arbitrary length solution - may take a while, but works on any positive number. public static final BigInteger factorial(final BigInteger n) { BigInteger factorial = BigInteger.ONE; for (BigInteger i = BigInteger.ONE; i.compareTo(n) <= 0; i = i.add(BigInteger.ONE)) { factorial = factorial.multiply(i); } return factorial; }
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
It is n! or n factorial.
P(n,r)=(n!)/(r!(n-r)!)This would give you the number of possible permutations.n factorial over r factorial times n minus r factorial
#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
Factorial for number N is N x N-1 x N-2 X N- (N-1). e.g. if you need to calculate factorial for 5 then compute 5 x 4 x 3 x 2 x 1.
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