A coin is flipped 14 times. How many different outcomes have at most 10 heads? 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?
 A: In fact, you did not follow the result from the other post. You're using permutations, rather than combinations. Instead, it should be $$2^{14}-\frac{14!}{14!0!}-\frac{14!}{13!1!}-\frac{14!}{12!2!}-\frac{14!}{11!3!}.$$
A: 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}$$14\choose{i}$.
which comes to 15914 if you use the formula for the ${n\choose k}={\frac{n!}{r!(n-r)!}}$.
A: Just use the binomial distribution to model the coin flips as $X\sim B(14,\frac12)$. Then the number of outcomes is $2^{14}$ so the number of required outcomes is given by
$$\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}$$
