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I know that there are a total of 4096 possible outcomes of tossing a coin twelve times, but I do not know how to calculate the number of possible outcomes with exactly 4 heads, with at least 2 heads, or at most 8 heads.

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Exactly 4 heads: This boils down to choosing 4 "spots" out of the 12 "spots" to put the 4 heads outcomes. That is, $\binom{12}{4}$.

At least 2 heads: $2^{12}- \binom{12}{0} - \binom{12}{1}$ subtracting the possibility of 0 or 1 heads.

At most 8 heads: $2^{12}- \binom{12}{12} - \binom{12}{11} - \binom{12}{10} - \binom{12}{9}$ subtracting the possibility of 9 or more heads.

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  • $\begingroup$ What do the two numbers in the parentheses mean? I haven't seen that before $\endgroup$ – Chris Dec 10 '13 at 4:52
  • $\begingroup$ @Craig en.wikipedia.org/wiki/Binomial_coefficient One way to calculate them is $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ where $n! = n \cdot (n-1) \cdot \dotsb \cdot 2 \cdot 1$. $\endgroup$ – Eric Auld Dec 10 '13 at 4:53

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