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How about you pay attention in class so you don't have to ask the computer for the answers.be smart listen to your teacher.

Just kidding ,i'm just thirteen what do i know

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Q: How do you simplify n factorial divided by n minus 1 factorial?
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What is a factorial function in Visual Basic?

' 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


Write a C-like program fragment that calculate the factorial function for argment 12 with do while loop?

#!/usr/bin/perl print factorial($ARGV[11]); sub factorial { my($num) = @_; if($num == 1) { return 1; # stop at 1, factorial doesn't multiply times zero } else { return $num * factorial($num - 1); # call factorial function recursively } }


Write a recursive procedure to compute the factorial of a number?

#include <iostream> using namespace std; int main() { int i, number=0, factorial=1; // User input must be an integer number between 1 and 10 while(number<1 number>10) { cout << "Enter integer number (1-10) = "; cin >> number; } // Calculate the factorial with a FOR loop for(i=1; i<=number; i++) { factorial = factorial*i; } // Output result cout << "Factorial = " << factorial << endl;


Programming to calculate a factorial number?

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


Write a program in java for factorial?

// 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; }