1 to 7 once?

Probability Level 4

All the 7-digit numbers containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once, and not divisible by 5, are arranged in the increasing order. Find the 2000-th number in this list.


The answer is 4315672.

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Aman Baser by Aman Baser , 21, India

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3 solutions

Aman Baser
Sep 7, 2015

The number of 7-digit numbers with 1 in the left most place and containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once is 6! = 720. But 120 of these end in 5 and hence are divisible by 5. Thus the number of 7-digit numbers with 1 in the left most place and containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once but not divisible by 5 is 600. Similarly the number of 7-digit numbers with 2 and 3 in the left most place and containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once but not divisible by 5 is also 600 each. These account for 1800 numbers. Hence 2000-th number must have 4 in the left most place. Again the number of such 7-digit numbers beginning with 41,42 and not divisible by 5 is 120 − 24 = 96 each and these account for 192 numbers. This shows that 2000-th number in the list must begin with 43.

The next 8 numbers in the list are: 4312567, 4312576, 4312657, 4312756, 4315267, 4315276, 4315627 and 4315672. Thus 2000-th number in the list is 4315672.

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I solved it basically the same way, but got the 600 as 5×5!=600. (As any number in the sequence is not divisible by 5, it cannot end in 5. Therefore, we can choose the last digit 5 ways and we can permutate the other 5 digits in 5! different ways.)

Similarly, I got 4×4!=96 and 3×3!=18 for the following digits.

Zee Ell - 5 years, 6 months ago
Razing Thunder
Jul 1, 2020 
1
2
 
from itertools import permutations
print([j for j in [int(''.join(list(i))) for i in permutations(['1','2','3','4','5','6','7'])] if j%5!=0][1999])

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Aayush Patni
Jan 13, 2016

Let. 1,2,3,4,5,6,7 be denoted by A,B,C,D,E,F,G respectively.

Then this question becomes simply like arranging words formed from letters A,B,C,D,E,F,G in a dictionary.

Some simple calculation leads to the 2000th word which is DCAEFGB.

Which is 4315672

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