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Question 191024: If all the six-digit numbers formed by using the digits 1,2,3,4,5, and 6, without repetition, are listed from least to greatest, which number will be 500th in the list?
Answer by solver91311(24713)   (Show Source):
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6! = 720 so we know that there are 720 different six digit numbers that can be formed with the digits 1 through 6 without repetition.
5! = 120, so there are 120 of these numbers that have a first digit of 1 and will be the smallest, then another 120 that start with 2 and will be the next larger group, and so on. All of the six digit numbers that have a first digit of 1, 2, 3, or 4 account for the first 480 ( 4 times 120 ) of the numbers in ascending numerical order. That means that the set that has a first digit of 5 accounts for the 481st through the 600th -- and that means the number we are looking for has a first digit of 5.
Continuing along this same logical path, the set of numbers that begin with 51 has 24 elements because 4! = 24. That means that the 481st through the 504th number in the list begins with 51 (480 plus 24 is 504).
3! = 6, so there are 6 of our numbers that start with 512, 6 that start with 513, and 6 that start with 514. 6 times 3 is 18, so the highest number that starts with 514 (namely 514632) is the 498th number in the list (480 plus 18 is 498). That means that the number we are looking for begins with 516.
516234 is then the 499th number in the list and 516243 is the 500th.
John
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