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About LinkBacksCount the ways that a 100 length binary string can be formed if the times that the digit zero appears is an even number and the times that the digit one (1) appears is an even number as well. (Note: in the even numbers zero is inluded.)
Dont get it.
Count the ways that a 100 length binary string can be formed if the times that the digit zero appears is an even number and the times that the digit one (1) appears is an even number as well. (Note: in the even numbers zero is inluded.)
Dont get it.
How many way can a binary string of length 100 be formed? (The binary string consists ONLY of 1's & 0's.)Originally Posted by milagros
Count the ways that a 100 length binary string can be formed if the times that the digit zero appears is an even number and the times that the digit one (1) appears is an even number as well. (Note: in the even numbers zero is inluded.)
Dont get it.
Then take half.
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?
You should say 'sixteen' ( since they are also sequenced ).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
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?
Obviously, 64.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
Maybe not so obvious is this:.
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?Count the ways that a 100 length binary string can be formed if the times that the digit zero appears is an even number and the times that the digit one (1) appears is an even number as well. (Note: in the even numbers zero is inluded.)
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
.
Thanks for being so analytical . Yesterday I thought exactly ( I thing ) the same. If I am correct you mean that my answer is 2^100 / 2 = 2^99 right?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?
You should say 'sixteen' ( since they are also sequenced ).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
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?
Obviously, 64.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
Maybe not so obvious is this:
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
.
I made it also with examples yesterday and I thing that I am right ?? And something important: I must express the result with a GENERATOR FUNCTION. Do
you know how to do this?
Thanks again.
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