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There are $32$ possible outcomes in total when a coin is tossed $5$ times. I have found that there are 10 possible outcomes that contain exactly$ 3$ heads by using $5C3=5!/3!2!$, but how do I find out how many possibilities contain at most $3$ heads? Thanks!

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  • $\begingroup$ Add up the number of possibilities that contain 1,2 or 3 heads. $\endgroup$ – Noe Blassel May 4 at 7:50
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    $\begingroup$ @NoeBlassel what about $0$ heads? $\endgroup$ – Mohammad Zuhair Khan May 4 at 7:54
  • $\begingroup$ Right, those too $\endgroup$ – Noe Blassel May 4 at 8:39
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This would be modeled with a sum of binomial coefficients. If you want precisely $k$ events to occur out of $n$ events, where event can only have success or failure, then the number of corresponding outcomes is $n$ choose $k$, i.e. $\binom n k$.

If you want at most some number of events, you can sum over the corresponding $k$ values, the $k$ values you deem valid. For example, if you want at most $3$ successes, you take the sum of $\binom n k$ for $k=0,1,2,3$.

In your case, $n=5$ and thus your result is

$$\sum_{k=0}^3 \binom 5 k$$

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It is $\displaystyle \frac{32}{2}+\binom 5 3$.

(Half of the possibilities are for getting $0$, $1$ or $2$ heads)

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Obviously there are $5$ possibilities for exactly $4$ heads (as there must occur exactly one tail) and only one for $5$ heads. Hence the number of possibilities for at most three heads equals $32-5-1$.

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