Codes to identify entries in a computer file are formed by using sequences of 7 ones and 5 zeros.?

Question

How many such sequences are possible?

Answer 1

That's not necessarily true.
Computers ulitmately work on binary code which is a combination of 0's and 1's, but there is no rule as to how many of each can appear in combination.

For example:
135=10000111

or

1548723=10000111101111010000110110011

Check out this URL:http://www.usbyte.com/common/binarysystem.htm

Answer 2

792

You have a sequence with 7 ones and 5 zeroes. Obviously, any such sequence must have a length of 12. The question then, is how many ways are there to position 5 zeroes in a sequence of 12 bits? If we consider each position as an object to be chosen, we recognize that it can be chosen at most once (you can't put two zeroes in the same position) and that it doesn't matter which order you place the zeroes in (if you place a zero the first, third, fifth, eleventh, and twelfth positions, you get the same sequence as if you placed a zero in the third, eleventh, fifth, first, and twelfth positions - namely 010101111100). Thus, the total number of such sequences is the number of combinations, without replacement, of five objects from a set of 12 - this is n!/(r!(n-r)!) which for this problem is 12!/(5!7!) or 792.

Answer 3

There is only one possible sequence of 7 ones and 5 zeros.

111111100000

Answer 4

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Answer 5

(2^7) 128 the combinations are cut down because not you are limited to 5 zeros and 7 ones. if you were asked the total out of twelve numbres regardless of numbers of 1's and 0's. that would be (2^12) 4096. out of a total of 4096 you only get 128 compinations of 7 ones and 5 ones.