A 6 digit number formed by 1,2,...6 such that no digit is repeated. What is the sum of all such numbers?

Total nos = 6*5*4*3*2*1 = 720 smallest = 123456 largest = 654321 average = (smallest + largest)/2 = 777777/2 = 388888.5 sum = avg * n = 388888.5 * 720 = 279999720 Hope the approach is right. Is it the right answer ?

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.