Sorry my post wasn't meaningful to you. A better example: How many ways can you have a string of 1's & 0's of length four? Code:

bit
cnt
0000 0 1
0001 1 2
0010 1 3
0011 0 4
0100 1 5
0101 0 6
0110 0 7
0111 1 8
1000 1 9
1001 0 10
1010 0 11
1011 1 12
1100 0 13
1101 1 14
1110 1 15
1111 0 16

You should say 'sixteen' ( since they are also sequenced ).

How many had an even number of 1's in the string.

If you said 'half' or 'eight' you are onto something.

Here is another example: How many ways can you have a string of 1's & 0's of length six? Code:

bit
cnt
000000 0 1
000001 1 2
000010 1 3
000011 0 4
000100 1 5
000101 0 6
000110 0 7
000111 1 8
001000 1 9
001001 0 10
001010 0 11
001011 1 12
001100 0 13
001101 1 14
001110 1 15
001111 0 16
010000 1 17
010001 0 18
010010 0 19
010011 1 20
010100 0 21
010101 1 22
010110 1 23
010111 0 24
011000 0 25
011001 1 26
011010 1 27
011011 0 28
011100 1 29
011101 0 30
011110 0 31
011111 1 32
100000 1 33
100001 0 34
100010 0 35
100011 1 36
100100 0 37
100101 1 38
100110 1 39
100111 0 40
101000 0 41
101001 1 42
101010 1 43
101011 0 44
101100 1 45
101101 0 46
101110 0 47
101111 1 48
110000 0 49
110001 1 50
110010 1 51
110011 0 52
110100 1 53
110101 0 54
110110 0 55
110111 1 56
111000 1 57
111001 0 58
111010 0 59
111011 1 60
111100 0 61
111101 1 62
111110 1 63
111111 0 64

Obviously, 64.

Maybe not so obvious is this: $\displaystyle 2^6 = 64 $.

How many had an

**even count** of 1's (or zeroes) in the string?

If you count them you will find 32.

Half of 64 (or 2^6) is 32 (or 2^5).

The FINAL EXAM!

How many ways can you have a string of 1's & 0's of length 100?

Recall:

a string length of 4 has 2^4 ways

a string length of 6 has 2^6 ways.

That should be easy.

The difficult part:

**Now take half of that.**
Recall: half of 2^6 is 2^5

.