Find the number of 17-digit binary sequences with more 0's than 1's.
What I know: If there are more 0's than 1's, the cases I have to calculate for is
9 0's and 8 1's
10 0's and 7 1's
11 0's and 6 1's
12 0's and 5 1's
13 0's and 4 1's
14 0's and 3 1's
15 0's and 2 1's
16 0's and 1 1's
17 0's and 0 1's
Is there other methods of doing this question? For example, using generating functions?
On
You are right that there are a couple of cases to be considered. You can have 9 zeros and 8 ones, which can be picked in $\binom{17}{9}$ ways. The next case is 10 zeros and 7 ones, which can be picked in $\binom{17}{10}$ ways. I guess you see the pattern. The total number of combinations with more zeros than ones is thus given by
\begin{align} \sum_{k = 9}^{17}\binom{17}{k}. \end{align}
Alternatively, you can use the approach of @5xum, who gave a hint at an easier way of solving this.
Hint: the number of binary sequences with more zeros than ones plus the number of binary sequences with more ones than zeroes plus the number of binary sequences with the same number of ones as zeros is...