We toss a coin $2n$ times: we want to know the probability $P(\#heads=\#tails=n)$. The number of possible arrangements is $2^{2n}$, the result is $P(\#heads=\#tails=n)=\frac{\binom{2n}{n}}{2^{2n}}$. I understand that$\binom{n}{k}$ is the number of possible set of size $k$ inside the set $n$. But can you explain why $\binom{2n}{n}$ is the number of favorable arrangements in this case?

And a similar problem, a set of $4n$ cards contains $2n$ black ones and $2n$ $red$ ones. We draw $2n$ cars and we want to know $P(\#red=\#black=n)$. Here we get $P(\#red=\#black=n)=\frac{\binom{2n}{n}*\binom{2n}{n}}{\binom{4n}{2n}}$; the question is the same as before why the above term is a binomial coefficient and why this times we multiply it?


$2n \choose n$ is the number of ways in which you can choose a set of $n$ objects from a set of $2n$ distinguishable objects.

This corresponds to the number of ways in which you get exactly $n$ heads (and therefore $n$ tails) with $2n$ tosses, since if you write down the outcomes of the tosses as a sequence (e.g. $HTTTHTH...$) then you are basically picking $n$ slots out of the $2n$ numbered the numbered slots. That is, in the example I just gave the heads occupy slots $1,5, 7...$, but another possible sequence would be THHHTTT..., and now they occupy slots $2,3,4$. So, how many possible sequences are there with exactly $n$ heads? As many different sets of $n$ numbers you can pick out of the numbers $1$ through $2n$, so that is $2n \choose n$

For the second question, first notice that the denominator (i.e. the 'below' term) is ${4n} \choose {2n}$, which corresponds to picking $2n$ cards out of $4n$ cards (all cards are different from each other).

Now, you need $n$ black cards, but since there are $2n$ black cards in the deck, you needs to pick $n$ out of $2n$ distinguishable black cards, so you have $2n \choose n$ options there. Likewise, you have $2n \choose n$ ways of picking $n$ red cards from the $2n$ red cards. And finally, you multiple, because for each 'choice' of black cards, you have all of the options of red cards.

A simpler example may demonstrate this latter principle. Say you have 4 red cards (label them $R1, R2, R3, R4$) and 4 black cards ($B1, B2, B3, B4$), and say you pick just $1$ red card and $1$ black card. Well, for each red card, I can add each of the $4$ black cards, so I have $4$ options for $R1$: $R1+B1$, $R1+B2$,$R1+B3$, and $R1+B4$. But you have those same $4$ options for the black card with any other red card, e,g, for $R2$ we have: $R2+B1$, $R2+B2$,$R2+B3$, and $R2+B4$. Another $4$ for $R3$, and another $4$ for $R4$, and you end up with 16 possible combinations of $1$ red card and $1$ black card, which is of course exactly $4*4$


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