human
Total number of non | Python3 program to count non - decreasing number with n digits.
dp [ i ] [ j ] contains total count of non decreasing numbers ending with digit i and of length j.
Fill table for non decreasing numbers of length 1. Base cases 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9.
Fill the table in bottom - up manner.
Compute total numbers of non decreasing numbers of length 'len  .
sum of all numbers of length of len - 1 in which last digit x is <= 'digit  .
There total nondecreasing numbers of length n won 't be dp[0][n] +  dp[1][n] ..+ dp[9][n].
Driver Code"
How can the above be solved in Python?

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What is a code to total number of non and the other steps required?
assistant
#Here is code to total number of non and the other steps required.
 def countNonDecreasing ( n ) :
	dp = [ [ 0 for i in range ( n + 1 ) ] for i in range ( 10 ) ]
	for i in range ( 10 ) :
		dp [ i ] [ 1 ] = 1
	for digit in range ( 10 ) :
		for len in range ( 2 , n + 1 ) :
			for x in range ( digit + 1 ) :
				dp [ digit ] [ len ] += dp [ x ] [ len - 1 ]
	count = 0
	for i in range ( 10 ) :
		count += dp [ i ] [ n ]
	return count
n = 3
print ( countNonDecreasing ( n ) )