ronnie1985 wrote:A 6 digit number formed by 1,2,...6 such that no digit is repeated. What is the sum of all such numbers?
Number of ways to arrange the 6 digits = 6! = 720.
Here's what the list of numbers looks like, from smallest to greatest:
123456,
123465,
123546....
654231,
654312,
654321.
Sum of the numbers in red = 123456+654321 = 777,777.
Sum of the numbers in green = 123465+654312 = 777,777.
Sum of the numbers in blue = 123546+654231 = 777,777.
As we proceed from the edges to the center, the sum of each pair = 777,777.
The number of pairs = 720/2 = 360.
Thus, the sum = 360(777,777) = 279,999,720.
An alternate approach:
The number of ways to arrange the 6 digits = 6! = 720.
Each digit will appear in each position an equal number of times.
Thus, the number of times that each digit will appear in each position = 720/6 = 120.
The sum of the digits 1 through 6, inclusive = 1+2+3+4+5+6 = 21.
Since each digit will appear in each position 120 times, the sum for each position in the number = 120*21 = 2520.
Each position in the number represents a power of 10.
Thus, the sum for each position (2520) must be multiplied all of the powers of 10 contained in the 6-digit integer:
2520 * (10� + 10� + 10³ + 10² + 10¹ + 10�) = 2520*111,111 = 279,999,720.
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