sum of all four digit numbers formed with digits 0,1,2,3,4 with repetition allowed.
Answers
The 4 digit number would take the form,
abcd, (where a,b,c,d are digits)
and, abcd = 1000a + 100b + 10c + d -----------(1)
Possible values of each of these digits:
elements(a) = 4,3,2,1
elements(b) = 4,3,2,1,0
elements(c) = 4,3,2,1,0
elements(d) = 4,3,2,1,0
To make pen-paper calculation simpler, let's fix values of a, and compute the remaining possible values of bcd.
Now, n(possible values of “bcd”) = 4P3 = 24 …. (Permutation: 3 slots to fill using 4 distinct digits) ------(2)
For a = 4, possible "bcd" values are
If we observe this pattern a little further, we’ll realize that 3 occurs exactly 6 times in hundreds place, exactly 6 times in tens place, and exactly 6 times in ones place.
We are to derive that, each of those digits (3,2,1,0) occur exactly 6 times in each of the 100s, 10s, 1s place. ----- (3)
So for a = 4, the sum of all 4 digits numbers of the form abcd, where b,c,d ∈ {3,2,1,0} such that b,c,d are not the same for any number abcd
= (24 X 1000 X 4)
+ 100(6 X (3 + 2 + 1 + 0))
+ 10(6 X (3 + 2 + 1 + 0))
+ 1(6 X (3 + 2 + 1 + 0))
---- here, overall formula comes from(1), 24 comes from (2) and 6 comes from (3),
Taking (3+2+1+0) as common, the formula can be further simplified to,
= (24 X 1000 X 4) + (6 X (3 + 2 + 1 + 0)) X (100 + 10 + 1)
= (24 X 1000 X 4) + (6 X (3 + 2 + 1 + 0)) X (111)
= (24 X 1000 X 4) + (666 X (3 + 2 + 1 + 0)) ------(4)
= 99,996
using (4)
For a = 3, Sum of all permissible numbers
= (24 X 1000 X 3) + (666 X (4 + 2 + 1 + 0)) = 76,662
For a = 2, Sum of all permissible numbers
= (24 X 1000 X 2) + (666 X (4 + 3 + 1 + 0)) = 53,328
For a = 1, Sum of all permissible numbers
= (24 X 1000 X 1) + (666 X (4 + 3 + 2 + 0)) = 29,994
Hence,
The sum of all 4 digit numbers that can be formed by the digits 0, 1, 2, 3, 4 (no repetitions)
= 99,996 + 76,662 + 53,328 + 29,994
= 259,980