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Python Program to Count Non-Bouncy numbers
• Last Updated : 20 Aug, 2020

If any number is represented in such a way that when we are reading it from left to right each ith Digit is greater or equal than i-1th digit is known as an increasing number. And if digits of any number are decreasing from left to right it’s known as decreasing number.

Example:

Increasing Number ?235668

all the digits from left to right are greater or equal to the previous digit.

Decreasing Number ? 653221

all the digits from left to right are lesser than or equal to the previous digit.

But if the number is neither increasing nor decreasing is Known as Bouncy Number.

Example:

523469 -> Some Digits from left to right are decreasing from left to right and some are increasing. So this is the example of Bouncy Number.

The task in this article is to count the total number of Non-Bouncy Numbers below 10k and print the final count in mod(109+7). To do this we will use the Stars and Bars method to calculate the number of non-bouncy numbers in the given range.

Stars and Bars Method:

Stars and Bars method is a technique that is used to deal with combination based problems. These type of problems arises when we want the number of identical groups.

The formula for calculating identical groups:

Where N are the identical objects and M is the container or range.

Final Formula:

Examples:

Input : k = 6
Output : 12951

Input : k = 9
Output : 140906


Below is the implementation:

Python3

 # import redunce function from functoolsfrom functools import reduce    # define a function to # calculate nCr def nCr(n, k):    # this approach is based on   # apporoach of stars and bar method  # using reduce and lambda function  # to calculate numer & denom  numer = reduce(lambda x, y: x * y,                 list(range(n, n - k, -1)))      denom = reduce(lambda x, y: x * y,                 list(range(1, k + 1)))              #  denom root of numer will be the final result  return numer // denom          # Driver Code# input value of kk = 6      # calculating r using function callr = int((nCr(k + 10, 10) +          nCr(k + 9, 9)          - 2 - 10 * k))  # print final resultprint(r % (1000000000 + 7))        

Output:

12951
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