Number of 10-digit binary strings with 6 ones, 4 zeros and no consecutive zeros The answer is $\binom{7}{4}$ but I can't figure out why.
 A: Place 6 ones in a row first:
.1.1.1.1.1.1.
where the .'s (there are 7) are places where we can put either a single 0 or nothing.
Note that 2 zeroes is not allowed (as a condition of the problem).
We have to put 4 zeroes on 7 candidate places, hence $\binom{7}{4}$.
A: Henno's answer gives a great combinatorics trick, but I wanted to share another one, which works nicely when you want to separate each zero by at least $n$ ones (and for this case $n = 1$).
First add $n$ more $1$s (which we'll get rid of later). Then make compound elements out of each zero, being $0$ followed by $n$ ones. Then count the arrangements of the compound elements and the remaining $1$s and for the actual resulting arrangement in any given case, disregard the last $n$ characters - which are all $1$s.
So for your example:
$$0,0,0,0,1,1,1,1,1,1 \\
\to 0,0,0,0,1,1,1,1,1,1,\color{red}1 \\
\to \color{blue}{01,01,01,01},1,1,1 \\
\text{ count arrangements: }{7 \choose 4} \\
$$
For an example with $15 \times 1, 4 \times 0, $ and requiring $0$s to be separated by at least three $1$s, 
$$
0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1 \\
\to 0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,\color{red}{1,1,1} \\
\to \color{blue}{0111,0111,0111,0111},1,1,1,1,1,1 \\
\text{ count arrangements: }{10 \choose 4} \\$$
A: Since we can't have consecutive $0$'s, any two of the four $0$'s are separated by at least one $1$.  By placing a $1$ between each pair of consecutive $0$'s, we obtain the string $0101010$.  We need to add three $1$'s to this string.  There are $5$ places in which to add the three $1$'s, namely before each of the zeros and after the last zero.  The number of ways to distribute three $1$'s into $5$ places is ${5+3-1 \choose 3} = {7 \choose 3}$. 
