Question

find the sum of all 4 digit numbers which can be formed from the digits 1, 2, 3 and 4 assuming that a digit can't be repeated in any one number.

okay... i know that there are 4! possible 4 digit numbers... but i have no idea how to get their sum O_O...

Answer 1

One way to do it is by making list of all the possible numbers and add them

Answer 2

1 way would be to find symmetry

see,,there will be 24 nos. in total as your rightly pointed out
also keep in mind any 4 digit no. = 1000a+100b + 10c +d

okay, so,
hope you understand that out of those 24 nos. there will be 6 nos starting from 1, 6 starting from 2, 6 from 3, and 6 from 4

same way, 6 nos. will be these with 1 in 2nd place, 6 with 2 in 2nd place, same for 3 and 4

when you add these nos.
you see a pattern :

(1000(1) + 100(1) + 10(1) +1 )*6 + (1000(2) + 100(2) + 10(2) +2 )*6 + ( ) *6 + ( )*6

hope you can fill those empty brackets

more simplification would yield you
(1000+100 +10 +1)(1+2+3+4)(6)

so ans = 66660

Answer 3

another method which i can think of is: ( and this is much shorter )

consider the smallest and largest no. formed from 1,2,3,4
they'd be 1234 and 4321 respectively
their sum = 5555

now 2nd smallest and largest nos. will be
1243 and 4312
their sum =5555

same way, 12 pairs of 5555 will be there when you add all
so final ans = 5555*12 = 66660

Answer 4

hope i could help .

Answer 5

Another method: