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i have this question but i don't know the answer please help me: How many bit strings of length 12 contain at least three 1 s and at least three 0 s?

gio
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  • @AndrewChin I sure hope you aren't suggesting that the answer be something like $2^6$ or $\binom{12}{3}\binom{9}{3}2^6$... If so, delete your comment because it is wrong. $2^6$ counts the number of bit strings of length $12$ where very specifically the first three bits are $1$ and the next three bits are $0$ and the remaining six bits are anything, so you clearly miss out on several possibilities. $\binom{12}{3}\binom{9}{3}2^6$ is far larger than $2^{12}$ which was the number of possible length 12 bitstrings with no restriction. – JMoravitz Feb 04 '20 at 16:57
  • Seems like it's not something that we even need to consider, thanks. – Andrew Chin Feb 04 '20 at 17:04

2 Answers2

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There are $2^{12}=4096$ bit strings altogeher. Of these, the number of bit strings containing exactly 0,1 or 2 1's is $${12\choose 0}+{12\choose 1}+{12\choose 2}=1+12+66=79$$Similarly, the number of bit strings containing exactly 0,1 or 2 0's is also 79. There is no overlap. Thenumber of bit strings satisfying the condition is 4096 - 2 $\times$79=4096-158=3938.

P. Lawrence
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The ways to place only two "ones" (remaining being "zeros") are $\binom {12}{2}$, and $\binom{12}{1}, \binom{12}{0}$ those of placing only one and none.
Same for the "zeros".

Taking out from the total $2^{12}= \binom {12}{0}+ \binom {12}{1}+ \cdots$ the above two tails gives the answer.

G Cab
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