How many 32-bit strings have fewer 1s than 0s? Now I know the answer is the summation of $32$ choose $0$ all the way to $32$ choose $15$. My class came up with this formula. I thought I understood how we got it but now I don't. If anyone would explain this method of shortening that calculation that would be great.
if $n$ is even $\ \ 2^{n-1 }  -$ $n-1 \choose n/2$.

Eric's answer is perfect, but let me just elaborate on what's going on.
Define a function $\tau$ on the set of 32-bit strings that swaps all $0$s for $1$s and vice-versa. So, for example,
$\tau(11100000000000000000000000000000) = 00011111111111111111111111111111$
$\tau(01010101010101010000000000000001) = 10101010101010101111111111111110$
Obviously $\tau$ will send all strings with $n$ digit $1$s to strings with $32-n$ digit $1$s, for all $n$. So you have three cases:


*

*Strings with between 0 and 15 digit 1s

*Strings with exactly 16 digit 1s

*Strings with between 17 and 32 digit 1s


$\tau$ swaps the first and third category with each other, and keeps the second category fixed (though swaps its members around). So the first and third categories must have the same number of elements. Also, the number of elements in the three categories add up to $2^{32}$; and the number of elements in the second category is $\binom{32}{16}$; so the number of elements in the first category is $\frac{2^{32} - \binom{32}{16}}{2}$.
