Hackety Hacking Problem Factorial: Difference between revisions

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== Curriculum Connections ==
== Curriculum Connections ==
== Solution ==
'''Ist Solution'''

f=1
puts("Enter the no. whose factorial to be calculated: ")
n=gets.chomp.to_i
def fact(n,k)
k=k*n
k
end
count=1
while count<n
f=fact(count+1,f)
end
puts f

Revision as of 03:35, 23 October 2007

Problem Statement

You have a word each made up of different letter no letter repeated.You have been given the length of the word.Now the problem is that you have to print the number of ways in which the letters can be arranged to return a new word. For Example: Word Length=3 take a word "bat" It can be written in 6 different ways: abt atb bat bta tab tba

So you have to write a program which takes the word length as the input from the user,And prints the number of ways in which they can be arranged.

Constrains

  • There can be more than one way to solve it.
  • You have to use recursion.
  • You have to give as many no.of solution as you can give

Help

Have you ever heard the word recursion, if not than let me explain. It is the name for algorithm which is to call the function from within itself but with changed parameters.

Sample Output

Enter the word length : 5
The number of ways the word can be arranged is = 120

Curriculum Connections

Solution

Ist Solution

f=1 puts("Enter the no. whose factorial to be calculated: ") n=gets.chomp.to_i def fact(n,k) k=k*n k end count=1 while count<n

f=fact(count+1,f) end puts f