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I followed the pattern here but it still resulted in my problem being incorrect. How many outcomes of a coin being flipped 12 times have exactly 4 heads?

(1 pt) A coin is tossed 14 times. d) How many different outcomes have at most 10 heads?

I did $2^{14}-\left(\frac{14!}{14!}+\frac{14!}{13!}+\frac{14!}{12!}+\frac{14!}{11!}\right)$, which translates to how many flips have at least $4$ tails.

Why isn't this working?

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  • $\begingroup$ If we flipped three times are $011$ and $110$ the same outcome or different? In other words, does order matter? $\endgroup$ – Alvin Lepik Apr 21 at 17:34
  • $\begingroup$ Can you explain where all you terms came from? What are all the factorial terms counting? $\endgroup$ – kccu Apr 21 at 17:35
  • $\begingroup$ @AlvinLepik I know that you use factorial when order matters but I couldn't find a way to do it with power either. $\endgroup$ – Adin D Apr 21 at 18:01
  • $\begingroup$ @kccu The factorial terms are the amount of flips with at least 4 tails. $\endgroup$ – Adin D Apr 21 at 18:02
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In fact, you did not follow the result from the other post. You're using permutations, rather than combinations. Instead, it should be $$2^{14}-\frac{14!}{14!0!}-\frac{14!}{13!1!}-\frac{14!}{12!2!}-\frac{14!}{11!3!}.$$

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  • $\begingroup$ Thank you. Question though. I thought I was suppose to use the formula c(n,r) -> n!/((n-r)!r!) for when order is important. Why is that not the case here? $\endgroup$ – Adin D Apr 21 at 18:05
  • $\begingroup$ That is for when order is not important, such as in this situation. When order matters, we use $$P(n,r)=\frac{n!}{(n-r)!}.$$ $\endgroup$ – Cameron Buie Apr 21 at 18:30
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Just use the binomial distribution to model the coin flips as $X\sim B(14,\frac12)$. Then the number of outcomes is $2^{14}$ so the number of required outcomes is given by $$\begin{align} 2^{14}\cdot P(X\le 10) &=2^{14}\cdot(1-P(X\ge 11))\\ &=2^{14}\cdot\left(1-\Bigg(\overbrace{\binom{14}{11}\left(\frac12\right)^{14}+\binom{14}{12}\left(\frac12\right)^{14}+\binom{14}{13}\left(\frac12\right)^{14}+\binom{14}{14}\left(\frac12\right)^{14}}^{P(X=11)+P(X=12)+P(X=13)+P(X=14)}\Bigg)\right)\\ &=2^{14}\cdot\left(1-\binom{14}{11}\cdot2^{-14}-\binom{14}{12}\cdot2^{-14}-\binom{14}{13}\cdot2^{-14}-\binom{14}{14}\cdot2^{-14}\right)\\ &=2^{14}-\binom{14}{11}-\binom{14}{12}-\binom{14}{13}-\binom{14}{14}\\ &=2^{14}-\frac{14!}{11!\cdot3!}-\frac{14!}{12!\cdot2!}-\frac{14!}{13!\cdot1!}-\frac{14!}{14!\cdot0!}\\ &=15914\\ \end{align}$$

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The number of ways of getting $0$ head is $14\choose{0}$. The number of ways of getting $1$ head is $14\choose1$.....The number of ways of getting $n$ head is $14\choose{n}$ with $1\le{n}\le{10}$.

so answer is $\sum_{i=0}^{10}$$14\choose{i}$.

which comes to 15914 if you use the formula for the ${n\choose k}={\frac{n!}{r!(n-r)!}}$.

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