# Python Challenge - Number Permutation

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Does anybody know how to do this? I’ve been trying, but I can’t.

from itertools import product as p def makeNumber(z): # Write your code here products = [] solutions = 0 for i in range(1,6): products.append(x for x in p(range(1,10), repeat=i)) for x in products: for y in x: if sum(y) == z: print(y) solutions += 1 return '\nThere are {} solutions'.format(solutions) print(makeNumber(3))

@aryabadri8979868753 found a solution, posted below

from itertools import product, chain def makeNumber(z): res = [] for i in range(1, 6): res.append([x for x in product((y for y in range(1, 10)), repeat=i) if sum(x) == z]) return len(list(chain.from_iterable(res))) print(makeNumber(45))

I did a brute force approach instead of recursion this time:
I iterated through all possible series of 5 numbers from 0 - 9 (with repetition), and counted only the relevant ones (meaning the ones where the numbers added up to the given number z, where there were no zeros that weren’t at the beginning).
I used an integer (meaning its digits) as each permutation, so that I could iterate through all the permutations.

my code
``````def checkPermutation(n, sum_of_digits = None, max_exp_of_10 = 4):
# get sum of digits of n, check if equals sum_of_digits
# can have leading 0s, but no 0s after a non-zero

previous_was_nonzero = False
total = 0  # will store sum of digits so far

checker = 10 ** max_exp_of_10

while checker >= 1:
digit = n // checker
if digit == 0:
if previous_was_nonzero:
# if zero comes after nonzero, return False
return False
else:
total += (digit % 10)
previous_was_nonzero = True
n = n % checker
checker = checker // 10

if sum_of_digits is None:
return (total == sum_of_digits)

def makeNumber(z):
if (z < 1) or (z > 45):
return 0

count = 0
for n in range(1, 100000):
if checkPermutation(n, z):
count += 1
return count

print(makeNumber(21))
``````
``````
def makeNumber(z):

lst = [[]] # temp lst to create tuple in, inner list.
cnt = 0 # store the count of permitations

for idx in range(5): # Limit to 5 digit in tuple
lst = [tuple(i,) + (x,) for i in lst for x in range(1, z+1) if x <= 9]
# create permutations in a lst

for i in lst: # loop trough list and put a condition to get the count of the sum that equals z
if sum(i) == z:
cnt += 1
return cnt

print(makeNumber(4))

``````