There is a rectangular table as shown in the figure, which has three chairs on two sides each. There married couples sit on these chairs.

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Find the number of ways in which exactly one couple sits in front of each other or adjacently.

What I tried:

Case $1$: The couple is sitting in front of each other.

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The number of ways: $$3\times\left(\binom31 \times 2\right)\times 2\times 2^2$$

$3$ is because the couple sitting in front can occupy any column, $\binom31$ is for choosing one pair of three, and $2$ is for interchanging the position of its members, the again $2$ is for the two diagonal pairs to swap positions and $2^2$ to interchange position of each member in a pair.

Case $2$: The couple is sitting adjacently.

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The number of ways: $$\binom41\times \binom31 \times 2\times 2^3$$

$4$ adjacent positions are possible,$\binom31$ is for selecting one out of three, $2$ is for interchanging the remaining couples, $2^3$ is for swapping position of members inside the pairs.

So total number of ways: $144+192=336$

But the given answer is $168$

The given solution is

The couple which sit infront of each other or adjacently can be selected in 3 ways and this couple when sit
Case-I: Infront of each other. Number of ways: = $3 × 2 × 2 × 2 = 24$
Case-II: Adjacently number of ways = $4 × 4 × 2 × 1 = 32$
Total number of ways so that exactly one couple sit infront of each other or adjacently
n(E) = $3(24 + 32) = 168$

  • 1
    $\begingroup$ You have over counted the adjacent pairs. It should be $4$ (number of adjacent pairs) times $\binom{3}{2}$ (how many ways you can make a couple) times $2$ (swapping position of the two couple members), which gives $24$. So $144+24=168$ $\endgroup$
    – Fotis
    Commented May 17, 2023 at 1:49
  • 1
    $\begingroup$ From the diagram, how are 4 & 6, for example, "adjacent" ? $\endgroup$ Commented May 17, 2023 at 5:40
  • $\begingroup$ Unless we are both misunderstanding something about the premise of the question, the given answer seems to be incorrect. I think your approach is mostly sound, except for one misstep in Case 1. Note that if the couple is sitting across from each other in the middle seats, the other two couples don't necessarily have to be diagonally across from each other, they could also be each on the same side of the table (since the couple in the middle would separate the other pairs.) This would increase the number of possibilities in Case 1 to 192. I believe the correct answer to this should be 384. $\endgroup$
    – A.J.
    Commented May 17, 2023 at 6:16
  • $\begingroup$ @Fotis But you haven't counted the cases in which members of a couple interchange position, no? $\endgroup$
    – Wolgwang
    Commented May 17, 2023 at 6:48
  • $\begingroup$ @A.J. trueblueanil Yeah I missed that. Thanks. $\endgroup$
    – Wolgwang
    Commented May 17, 2023 at 6:49

1 Answer 1


There are some errors in both the reasoning of the OP and that of the textbook. The textbook's solution is comically tragic : neither does it give details, nor does it get things right.

Let's understand the OP's understanding line-by-line. (Fantastic diagrams : thank you!)

The conditions of the question is that exactly one couple must be adjacent to each other or opposite to each other. The idea of the OP is to fix this couple's position first and then think about the other positions. Obviously, the "chosen" couple is either adjacent to each other or opposite to each other, so the break up into cases is logical, because the two cases are disjoint. So far, so good.

Now let's look at Case 1. The OP's reasoning is incorrect, as explained in the comments. Here's why.

Suppose the couple is opposite each other and sitting in positions $1$ and $4$. Which couple this is, is chosen in three (using English words here to avoid confusion with the position numbers) ways. Then who sits on $1$ and who sits on $4$ is decided in two ways. At this point, consider the person sitting on position $2$. It must happen that their spouse must sit on position $6$, because they cannot be in position $3$ (adjacent) or in position $5$ (opposite). Now the other couple must go in positions $3$ and $5$. This is pretty much the reasoning of the OP and is spot on. You can choose the person sitting in position $2$ in four ways, and the arrangement of the other couple between positions $3$ and $5$ in two ways. All in all, that yields $3 \times 2 \times 4 \times 2 = 48$ ways.

The above argument works even when the couple who are sat opposite each other are sat in positions $3$ and $6$, because of symmetry : any such configuration can be rotated by 180 degrees to reveal another configuration satisfying the same properties but where $1$ and $4$ are the positions of the opposite couple. Make that another $48$ ways.

However, the argument fails when the couple sat opposite each other are in positions $2$ and $5$. The reason is clear : if someone sits in position $1$, then their spouse can sit in position $3$ now, because it's not adjacent to position $1$ anymore. Here's the point until where the argument is correct : the couple can still be chosen in three ways, and their positions among $2$ and $5$ can still be chosen in two ways. Now, fix the person in position $1$. Their spouse can be sat in positions $3$ or $6$. In each case, the other couple occupies positions $4,6$ and $3,4$ respectively, both of which are acceptable. So there are four ways to fix the person sitting in position $1$, then two ways to select whether their spouse sits in position $3$ or $6$, and then two ways to decide how the last couple sits in positions $3,6$ or $4,6$ depending upon the choice of the spouse of the person in position $1$. All in all, we have $3 \times 2 \times 4 \times 2 \times 2 = 96$ ways of couples satisfying this criterion.

All in all, Case I entails $48+48+96=192$ possibilities, not $144$.

Looking at Case 2, the reason why the OP's reasoning is correct is because regardless of the position of the adjacent couple, the counting process is the same. For example, if the adjacent couple is positioned at $1,2$, then flipping the top and bottom occupants lead to the adjacent couple being at $4,5$. Now rotate this by $180$ degrees and the adjacent couple is sitting at $2,3$. It should be quite clear that it is enough to argue the case when the adjacent couple is at positions $1,2$. That's why the diagram of the OP accurately represents the situation.

Their logic for Case 2 is correct as well. Indeed, once the adjacent couple in $1,2$ is positioned, then we think about the person in position $5$. Their spouse must be in position $3$ and the other two must be in spots $4,6$, so that gives $4 \times 2$ ways. Accounting for the choice of the adjacent couple and their position, another $3 \times 2$ comes in. Accounting for the fact that we fixed positions $1,2$ , another $4$ needs to be multiplied. All in all, that is $4\ times 4 \times 2 \times 2 \times 3 = 192$ ways.

This totals up to $384$ ways, which is the correct answer.

To provide comic relief, let's talk about how bad the book is at doing its job. I think it starts off well by deciding to fix the couple sitting adjacent to/opposite each other. It just takes a $3$ out of the calculations. However, after that, what occurs is abysmal.

Take the in-front case. First of all, the multiplication by $3$ in the in-front case : we already discussed why this is wrong. But even if you fix the in-front couple to sit in , say $1$ and $4$, it is impossible to get only four more ways to position the couples. $8$ is just the number of ways you can exchange the positions of the spouses, so how the author forgets this much is really bizarre.

The other case is also off by $2$. I understand the first $4$ comes from the fixing of the positions $1,2$ for the adjacent couple (probably!) and the second $4$ probably comes from fixing the person sitting in $5$, with the $2$ coming from switching the couple sitting in $4$ and $6$. However, this case fails to take into account the fact that the couple in $1,2$ can be switched with each other. That means it misses the mark by a factor of $2$.

So that's where both of you go wrong, and the right solution is $384$.

  • $\begingroup$ This was from FIITJEE AITS XD .Thanks a ton! By any chance, do you remember me? I have JEEA in 16 days :) $\endgroup$
    – Wolgwang
    Commented May 18, 2023 at 13:49
  • 1
    $\begingroup$ @Wolgwang Of course! It's been a long time, and I thought I should write a nice SV-type answer after some time. Good luck for JEEA : I hope that a prolonged stay on MSE helps you ace the math section of the exam. For what it's worth, MSE was not useful when I was preparing for JEEA (this was 2013, the very first year that JEEA was introduced). At that point, most people would actually recommend spending time on very carefully chosen questions first (generally, combinatorics questions are easier than others, for example) and then move to the hard ones (or skip them if there's no time). $\endgroup$ Commented May 18, 2023 at 23:25
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    $\begingroup$ Now, thankfully the variety of answers on MSE has gone up, so all the little tricks and patterns are out in the open. That makes it easier for you to get through, so I really hope that you do well. $\endgroup$ Commented May 18, 2023 at 23:26
  • $\begingroup$ @Wolgwang I hope your exam went well. $\endgroup$ Commented Jun 13, 2023 at 5:23
  • 1
    $\begingroup$ Not really :( Have taken admission in a state university. $\endgroup$
    – Wolgwang
    Commented Jun 29, 2023 at 11:22

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