If I'm counting the number of binary strings of a certain length with a certain number of 1's, should I use combinations or permutations? And should I use repetition allowed, or repetition not allowed formula?
A binary string is a string with 1's and 0's in a row. {0,1} is a different string from {1,0}. Say I'm considering binary strings of length 6 and I want to count the number of strings with 3 1s (and 3 0s). Should this be regarded as a combination or permutation problem? And is this considered to be "repetition allowed?" It seems like it's "repetition required." 
 A: You can often solve a counting problem by working out in how many ways you can choose the required object.  When doing so, there are a number of questions that are nearly always worth asking yourself.


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*What am I choosing?  In this example, are you choosing the $1$ or the $0$s?  No - you know all about these, there are three $1$s and three $0$s.  What you are choosing is the places in which the $1$s and $0$s will occur.  In this case, once you have chosen the places for the $1$s you know where the $0$s go (namely, in all the remaining places), so we only have to choose places for the $1$s.

*So, you have to choose three places from six possibilities.  Is repetition allowed?  If you are not sure, make up a specific instance and ask whether or not it makes sense.  So, could we put three $1$s in places $6,6$ and $6$?  Obviously not - the three $1$s must go in different places.  Note how important question 1 is - if you thought you were choosing the $1$s (rather than the places) you would certainly say that repetition is allowed because there are three of them.  This would be wrong.

*Is order important?  Once again, if you are not sure, ask yourself about a specific instance.  For example, if I choose to put $1$s in places $1,2$ and $4$, is that the same or different from putting them in places $2,4$ and $1$?  I hope it is clear that it is the same (the word is $110100$ in both cases) and therefore order is not important.


So, we have to choose $3$ places from $6$, with repetition not allowed and order not important, and the number of ways to do this is $C(6,3)$.
