Question 1101136
<pre>
If we consider having the non-natural number 0000000 along
with the natural numbers in binary from 0000001 through 1111111, 
along with all the natural numbers in a list, like this:

0000000
0000001
0000010
0000011
0000100
0000101
0000110
0000111
...
1111110
1111111

Each column of digits in that list contains
the same number of 1's as 0's

The last binary number in that list, 1111111, has decimal
value

1(2^6)+1(2^5)+1(2^4)+1(2^3)+1(2^2)+1(2^1)+1(2^0) =

1(64)+1(32)+1(16)+1(8)+1(4)+1(2)+1(1) =

64+32+16+8+4+2+1 = 127, 

so counting 0000000 there are 128 binary numbers in that list.

Each column of digits contains 128 digits, 64 or which 
are 0's and 64 of which are 1's.  Since there are 7 
columns of digits, there are 7(64) = 448 0's and 448 1's.

Edwin</pre>