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?
Best Answer
In fact, you did not follow the result from the other post. You're using permutations, rather than combinations. Instead, it should be $$2^{14}-\frac{14!}{14!0!}-\frac{14!}{13!1!}-\frac{14!}{12!2!}-\frac{14!}{11!3!}.$$