How many different bit strings of length 10 contain at most eight 1's?

Question

How many different bit strings of length 10 contain at most eight 1's?

My work

$C(10,0)$+$C(10,1)$+$C(10,2)$+$C(10,3)$+$C(10,4)$+$C(10,5)$+$C(10,6)$+$C(10,7)$+$C(10,8)$

Answer 1

For the sake of removing this from unanswered queue:

Yes, your answer is correct, albeit tedious.

A more concise way of finding and writing the answer would be to count the complementary event. There are $\binom{10}{10}+\binom{10}{9}=1+10=11$ ways to get a string of length ten with strictly more than eight 1's.

Subtracting this from $2^{10}=1024$, the total number of length ten binary strings with no other restrictions, gives us the total number of length ten binary strings with eight or fewer 1's as:

$$2^{10}-\binom{10}{10}-\binom{10}{9}=1024-11=1013$$

I would not have wanted to try adding $\binom{10}{0}+\binom{10}{1}+\binom{10}{2}+\dots+\binom{10}{8}$ manually in a direct fashion, but using the above the arithmetic was quite easy to work with.