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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 ■
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
It is n! or n factorial.
n! = 1*2* ... * (n-1)*n
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.
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 ■
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
It is n factorial, written as n!
1 is one.
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); }
First, the answer to what is factorial. Factorial is denoted by '!' N! = N * (N-1) * (N-2) * ... * 3 * 2 * 1. For example, 5! = 5 * 4 * 3 * 2 * 1 = 120. There are two approaches to coding a function that will calculate n!. One does it iteratively, and the other recursively. Iteratively: factorial (n){ result <- 1 for(i <- 1; i <= n; i <- i + 1){ result <- result * i } return result } Recursively: factorial (n){ if n is 1 return 1 return n * factorial(n - 1) } Note that n must be 1 or greater for each of the above methods.
' 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; }
#include<stdio.h> #include<conio.h> main() { int f=1,i=1,n; clrscr(); printf("\n Enter factorial value"); scanf("%d",&n); for(;i<=n;i++) { f=f*i; } printf("\n The factorial value=%d",f); getch(); }