How many bit strings of length 5 are thereSolutionEvery bit

How many bit strings of length 5 are there?

Solution

Every bit can either be a 0 or 1. So to find the amount of bit strings of length either, you do 2length to find the amount of bit strings there are of a given length.


There are 2^n bit strings of length n, so..
\"6 or less\" = length-6, + length-5 + ... + length-1, which is
2^6 + 2^5 + 2^4 + ... + 2, which is
64 + 32 + ... + 2, which is
126. (It would be 127 if you count \"bit strings of length 0\" but probably that is not wanted :-) )

The next one is more difficult because the easy way to calculate it is not obvious. There are 26^4 strings of four lowercase letters. Of those, 25^4 contain no \"x\" (as they are strings of length four containing only members of the 25 remaining letters). So the answer of those which contain (at least one) x is the difference between those numbers: 26^4 - 25^4 = 456,976 - 390,625 = 66,351.

Each different function would map a digit from the first set to either 1/0, which is two different options per digit. So there are 2^n such functions, for a domain of n numbers. (e.g., for 3 numbers, 123 could map to 000, 001, 010, 011, 100, 101, 110, 111, respectively = 2^3 functions)

The number of orderings for a set of 5 things: there are 5 different runners who could finish first. Eliminating that one, there are 4 left who could finish second. That leaves 3 who could finish third, etc. 5*4*3*2*1 = 5!. The answer is 5! which is 120.

The \"17 out of 40\" problem is 40C17, or (40!)/(17! * (40-17)!). Think of it this way: There are 40! different orderings of 40 questions. But 17 of those are true and there are 17! ordering of those true ones which do not change the answer key. Also there are 23 (40-17) which are false and there are 23! orderings of the false questions that do not change the answer key. So dividing out the alternatives that don\'t change the answer key, leaves about 88,732,379,000 different answer keys.
Edited 3 months ago