This is a homework problem. I know what the answer is, but I want to understand it.

At a party, $15$ married couples are seated at random at a round table. Suppose that of these married couples, $5$ husbands and their wives are older than fifty and the remaining husbands and wives are all younger than fifty. What is the probability that all men over fifty are sitting next to their wives?

My attempt at a solution:

We want to seat $30$ people at a round table. That means we have $\frac{30!}{30}=29!$ ways of seating them. Let's define the event $A$ as the event that the 5 elderly couples are seated next to each other. We have $5$ married couples older than fifty and $10$ other couples. Since the order of the other couples is of no significance we can consider them to be 20 random distinguishable people. The first married couple over fifty can sit in any adjoining chairs and since there is a rotation symmetry they can only do that in one way since we will define the seats they sit in as chair $1$ and chair $2$. The second couple can sit in any of the $28$ chairs left, the third can sit in any of the $26$ chairs left, etc. After seating the couples, we have 20 people left who can be seated in $20!$ ways. Every elderly couple can switch places giving a new seating arrangement that also guarantees they sit next to each other so that gives us another $2^5$ ways. That makes the total of ways $N = 28 \cdot 26 \cdot 24 \cdot 22 \cdot 20! \cdot 2^5$. So $p(A) = \frac{N}{29!}= \frac{28 \cdot 26 \cdot 24 \cdot 22 \cdot 20! \cdot 2^5}{29!}$

This turns out to be incorrect, the answer should be: $p(A)=\frac{24! \cdot 2^5}{29!}$.

Can anyone help me understand the fallacy in my argument and help me understand the correct solution?


2 Answers 2


We have five couples that need to be seated next to one another. So for now, let us think of each of these couples as a single person. This means that we have twenty five people. The number of ways to seat them at a round table is $24!$. Now, for every elderly couple we have two different ways to seat them wife next to husband, and hence the total number of ways is $24!\cdot2^5$. Dividing by $29!$ yields the answer.


You can think of your event A in another way.
Since an older man has to sit next to his wife, we take both older wife and husband as one unit. Thus, you will have 20 random people + 5 units(older couples). So total you have 25 persons which can be seated in $ \frac{25!}{25}$ ways. In addition to that, an older husband and his wife can be seated next to each other on two ways: first wife then husband OR first husband then wife.
So, your answer will be $ \frac{24!.2^5}{29!}$.


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