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Each married couple as well as each dancing pair consists of a man and a woman. How many possible pairings are there?

Here is the same question with a different amount of couples. I read the answers to the linked question but still couldn't get this right.

This question was asked in a TV show. It was part of the following problem:

Find the name of a car manufacturer. Its second letter is A. Don't guess, but find out the name in the following way: Count the number of pairings [for the problem above]. Subtract 814501449252729612547 from the answer. Split the resulting number into sequences of length 2. Interpret these sequences as ASCII codes and you will get the letters in the name of the manufacturer. Rearrange the letters and you will find the answer.

I have hidden the answer just in case someone wants to figure it out by themselves. You can see the answer by placing the mouse cursor over the next line.

SALAMANCA

I tried to use the inclusion-exclusion principle as in one of the answers in the linked question. That is how I came up with this:

$$ \begin{equation*} \begin{split} & {25 \choose 7} \left[18! - \sum_{i=1}^{18} {18 \choose i} (18-i)! (-1)^{i+1} \right]-814501449252729612547 \\ & = 317692059206291594553 \end{split} \end{equation*} $$

The first part, ${25 \choose 7}$, is the number of ways to choose 7 married couples. $18!$ is the number of all pairings for the remaining 18 couples. The sum over $i$ is the number of pairings where at least one pair is a married couple.

Here it is in WolframAlpha.

However the number does not make sense as an ASCII code sequence and it is not even close to the correct answer.

EDIT: I added one dash that I had forgotten to the formula, on the WolframAlpha page too. The answer still doesn't contain the correct ASCII code points.

I also tried computing derangements (WolframAlpha):

$$ \DeclareMathOperator*{\round}{Round} \begin{equation*} \begin{split} & {25 \choose 7} \round\left(\frac{18!}{e}\right) - 814501449252729612547 \\ & = 317692059206291594553 \end{split} \end{equation*} $$

where $\round$ is the nearest integer function. The answer is the same one.

Community
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    For the dancing people the unlucky couples can be chosen in $\binom{25}{7}$ ways. Multiply by the number of derangements of a set of $18$. That number can be computed by Inclusion/Exclusion, please see derangements in Wikipedia. There is a simpler formula in terms of $e$. – André Nicolas Jul 20 '15 at 21:46
  • Thanks! I tried this and edited the question. –  Jul 21 '15 at 07:57

1 Answers1

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First we need to select $7$ men, that will dance with their own wives. It can be done in ${25 \choose 7}$ ways. Now we have $18$ men willing not to dance with their own wife.

In other words - we need to permutate $18$ men without constant points, ie. derearrange them. It can be done in $!18$ ways.

  • $!1=0$, $!2=1$
  • $!n = (n-1)(!(n-1)+!(n-2))\,$ for $n>2$

The answer is then $${25 \choose 7}\cdot !18$$

Edit

Well, now I've red the whole question and comments and see, that it was already answered and solved. By the way it was fun solving it and recalling some informations not used for years :)

Jaroslaw Matlak
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