Each permutation of the numbers 1,2,3,4,5,6,gives a six-digit number. If the number corresponding to all possible permutations are listed in the increasing order of magnitude, find the 319th number on the list. Find the sum of all the numbers formed by all the permutations.
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The permutations starting with 1 number 120 in all.
The permutations starting with 2 number 120 in all.
The permutations starting with 31 number 24 in all.
The permutations starting with 32 number 24 in all.
The permutations starting with 34 number 24 in all.
The permutations starting with 351 number 6 in all.
This is a total of 318 permutations.
Consequently, the 319th is: 352146
A list of all permutations consists of 720 numbers, each with six digits. To add these number, consider that each of the ones', tens', etc. places contain each of the digits 1, 2, 3, 4, 5, 6 a total of 120 times. Thus, the sum for each column is: (1+2+3+4+5+6)*120 = 2520. Consequently, the total of all the permutations is:
2520+ 25200+ 252000+ 2520000+ 25200000+ 252000000 = 279999720
The permutations starting with 2 number 120 in all.
The permutations starting with 31 number 24 in all.
The permutations starting with 32 number 24 in all.
The permutations starting with 34 number 24 in all.
The permutations starting with 351 number 6 in all.
This is a total of 318 permutations.
Consequently, the 319th is: 352146
A list of all permutations consists of 720 numbers, each with six digits. To add these number, consider that each of the ones', tens', etc. places contain each of the digits 1, 2, 3, 4, 5, 6 a total of 120 times. Thus, the sum for each column is: (1+2+3+4+5+6)*120 = 2520. Consequently, the total of all the permutations is:
2520+ 25200+ 252000+ 2520000+ 25200000+ 252000000 = 279999720
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Starting with 1 there are 5!=120 different numbers
starting with 2 there are 5!=120 different numbers
starting with 3 there are 5!=120 making a total of 360 different numbers.
The 319th number must start with 3 and you should be able to
find this number .I make it 325614.
There are 6!=720 perms and each digit in each position occurs 120 times.
Sum of the 6 digits=21so sum of all numbers
= (120X21)(10^5+10^4+10^3+10^2+10+1)=120X2…
=279999720, I think.
starting with 2 there are 5!=120 different numbers
starting with 3 there are 5!=120 making a total of 360 different numbers.
The 319th number must start with 3 and you should be able to
find this number .I make it 325614.
There are 6!=720 perms and each digit in each position occurs 120 times.
Sum of the 6 digits=21so sum of all numbers
= (120X21)(10^5+10^4+10^3+10^2+10+1)=120X2…
=279999720, I think.