[ Combinations and Permutations ]

In the final of a play competition for March 8, two plays made it to the final. In the first play, $n$ students from class 5A participated, and in the second play, $n$ students from class 5B participated. At the play, $2n$ mothers of all $2n$ students were present. The best play is chosen by the mothers' votes. It is known that exactly half of the mothers vote honestly for the best play, while the other half always vote for the play in which their child participates.

a) Find the probability that the best play will win by a margin of votes.

b) The same question, if more than two plays made it to the final.

#
a) Let's call a mother honest if she votes for the best play in any case. It is known that $n$ honest mothers will vote for the best play, so it will not lose for sure (half of the votes are definitely for it). The only case when the best play does not win is a tie, when the worst play also gets half of the votes. This is only possible if all the children of dishonest mothers are in the worst play. Let's find the probability of this event.

There are $C_{2 n}^{n}$ cases of distributing $n$ dishonest mothers among the total number of $2 n$ mothers. And only in one of these cases do all dishonest mothers vote for the worst play. Thus, the probability of a tie is $\frac{1}{C_{2 n}^{n}}=\frac{(n!)^{2}}{(2 n)!}$. Therefore, the probability of the event "the best play will get more votes" is $1-\frac{1}{C_{2 n}^{n}}=\frac{(2 n)!-(n!)^{2}}{(2 n)!}$

b) Let there be $m>2$ plays. In this case, the best play wins for sure - it gets at least half of the votes, one from each honest mother. The votes of the other mothers are divided among the other plays, each of which gets no more than $1 /{ }^{1}$ of all votes, which is less than half.

## Answer

a) $\frac{(2 n)!-(n!)^{2}}{(2 n)!}$. b) 1.