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Permutation_Problem.py

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"""

How many different ways can you arrange 3 digits to find the product of a 2-digit number and a 1-digit number?

"""

# find the smallest 3 digit number that is a multiple of a two digit number and an one-digit number

# it is 100 (= 2 x 50)

# and maximum is 891 (=9*99)

# run a loop from 100 to 891 to find the other 3 digits number between them, which are multiples of ...

multiples = []

for i in range(100,892):

for j in range(2,10):

if i % j == 0:

multiples.append(i)

break

# print(multiples) # these multiples are the possible 3-digit numbers

print(f'Total {len(multiples)} multiples are available.')

# now we have to find number of ways, each digit place can be arranged

digit = {}

for number in multiples:

for pos,digits in enumerate(str(number)):

if digits in digit.keys():

digit[digits] = digit[digits] + 1

else:

digit[digits] = 1

print(digit)

# thus our answer should be 610