Given two positive integers n and r where n > r >1. The task is to find the value of f(n)/(f(r)*f(n-r)). F(n) is deined as follows:
1-1 *2-2 *3-3 *….. n-n
Input: n = 5, r = 3 Output: 1/200000 Input: n = 3, r = 2 Output: 1/27
A naive approach to solve this question is to calculate f(n), f(r) and f(n-r) separately and then calculating the result as per given formula but that will cost a bit high of time complexity.
A better approach to solve this question is to find the greater value among r and n-r and then after using the property f(n) = f(n-1)* n-n = f(n-1)/nn of given function, eliminate the greater among f(r) and f(n-r) from numerator and denominator. After that calculate the rest of value by using simple loop and power function.
find max(r, n-r).
iterate from max(r, n-r) to n
result = ((result * i-i / (i-max(r, n-r)) -(i-max(r, n-r)) )
Below is the implementation of the above approach:
- Find 2^(2^A) % B
- Find value of (1^n + 2^n + 3^n + 4^n ) mod 5
- Find value of (n^1 + n^2 + n^3 + n^4) mod 5 for given n
- Find (1^n + 2^n + 3^n + 4^n) mod 5 | Set 2
- Find the value of max(f(x)) - min(f(x)) for a given F(x)
- Find (a^b)%m where 'b' is very large
- Find the other number when LCM and HCF given
- Find minimum x such that (x % k) * (x / k) == n | Set-2
- Find minimum x such that (x % k) * (x / k) == n
- Find two numbers whose sum and GCD are given
- Find x and y satisfying ax + by = n
- Find x, y, z that satisfy 2/n = 1/x + 1/y + 1/z
- Given two numbers a and b find all x such that a % x = b
- Find two integers A and B such that A ^ N = A + N and B ^ N = B + N
- Find the ln(X) and log10X with the help of expansion
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to firstname.lastname@example.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.