There is a bit string of length $70$. At least one of the following restrictions must apply:
i)The first $9$ bits cannot contain exactly $5$ 1s
ii)The first $49$ bits cannot contain exactly $27$ 1s.
How many combinations are there?
My first instinct is to do the following: $2^{70}$ and subtract some number of combinations.
I am not sure how to go about this or if there is a more elegant way. I am currently getting the following result, but I am not confident in it because I believe that there may be some double counting:
$2^{70}-(\binom{9}{5}+\binom{49}{27})=$
$997,056,092,945,595,000,000$
Your instinct is good. However, you have not counted the bad cases correctly.
As you observed, there are $2^{70}$ possible bit strings. From these, we must subtract those bit strings in which the first $9$ bits contain exactly $5$ 1s or those in which the first $49$ bits contain exactly $27$ ones.
Bit strings of length $70$ in which the first $9$ bits contain exactly $5$ ones: There are $\binom{9}{5}$ ways to select which $5$ of the first $9$ bits are filled with 1s. There are $2^{61}$ ways to fill the remaining $61$ bits. Hence, there are $$\binom{9}{5}2^{61}$$ such bit strings.
Bit strings of length $70$ in which the first $49$ bits contain exactly $27$ ones: There are $\binom{49}{27}$ ways to select which $27$ of the first $49$ bits are filled with 1s. There are $2^{21}$ ways to fill the remaining $21$ bits. Hence, there are $$\binom{49}{27}2^{21}$$ such bit strings.
Notice that if we subtract these numbers from the total, we will have subtracted those cases in which exactly $5$ of the first $9$ bits and exactly $27$ of the first $49$ bits are 1s twice. We only want to subtract such cases once, so we need to add them back.
Bit strings of length $70$ in which exactly $5$ of the first $9$ bits and exactly $27$ of the first $49$ bits are 1s: There are $\binom{9}{5}$ ways to select which $5$ of the first $9$ bits are filled with 1s. If there are also exactly $27$ 1s in the first $49$ bits, then $27 - 5 = 22$ of the next $49 - 9 = 40$ must be filled with 1s. The remaining $21$ bits can be filled in $2^{21}$ ways. Hence, there are $$\binom{9}{5}\binom{40}{22}2^{21}$$ such cases.
Hence, by the Inclusion-Exclusion Principle, the number of admissible bit strings is $$2^{70} - \binom{9}{5}2^{61} - \binom{49}{27}2^{21} + \binom{9}{5}\binom{40}{22}2^{21}$$