Probability of friends sitting next to each other on a plane?

Question

A small plane has 12 seats, 2 rows with 3 people on each side. Four friends, A, B, C and D, are flying with the plane. Seats are selected at random for everyone on the plane. What is the probability that at least two friends sit next to each other? Two adjacent isle seats does not count

I can not seem to find any good apporach to this problem. My first apporach was to order the elements in a long row where the 1st, 2nd and 3rd place correspond to the first 3 seats to the left and so on.

I started with finding the number of ways I can order 12 elements: $12!$. Then I tried to count the number of elements that have at least two people next to each other by doing this: $\frac{4!}{2!\cdot8!}$. However, I realize that this of course counts duplicates and I need to remove these duplicates but I do not see how. After this I also need to remove the permutations where two friends are next to each other on the 3rd and 4th place, 6th and 7th place and 9th and 10th place, but I can not find a easy way to do this either.

Answer 1

As Daniel said in the comments, this can be easily solved by considering the case where no friends sit together.
Case 1: All friends sit in different sections. No. of ways to distribute passengers among sections $= 4!$. No. of ways a passenger can sit in his section $=3$. Total cases:$$4!3^4$$ Case 2: Two friends sit in one section and the other two sit in different sections. Ways to pair up two friends $= {4\choose 2}$. Ways to assign sections $= {4\choose3}3!$. The two people sitting alone each have 3 choices but the two people sitting together only have 2 choices. So no. of cases:$${4\choose{2}}{4\choose3}3!3^22$$ Case 3: Two pairs of two friends each. Ways to assign pairs=${\frac{4\choose2}{2!}}$. Ways to assign sections$={4\choose2}2!$. Ways one pair can sit in their seats$=2$. No of cases:$${4\choose2}{4\choose2}2^2$$No of total unfavourable cases evaluates to sum of these$=4680$. No of total cases $=12\times11\times10\times9=11880$. Probability of no friends sitting together$$\frac{4680}{11880}=\frac{13}{33}$$ Probability of at least two friends sitting together$$1-\frac{13}{33}=\frac{20}{33}$$ EDIT: corrected mistake pointed out by user2661923

Answer 2

We will start by computing the complementary probability--the probability that no two friends are in adjacent seats. There are $\binom{12}{4}$ ways to choose four seats, all of which we assume are equally likely. We want to count the number of those in which no two seats are adjacent.

We ask, more generally, what are the number of ways to select $r$ seats, no two adjacent. Consider one half-row of three seats. There are three ways to place one person, and only one way to place two persons; so the generating function for the number of ways to place people in the three seats is $1 + 3x + x^2$. The generating function for the number of ways to place $r$ persons in four half-rows is then $$\begin {align} f(x) &= (1 + 3x +x^2)^4 \\ &= 1+12 x+58 x^2+144 x^3+\color{red}{195} x^4+144 x^5+58 x^6+12 x^7+x^8 \end{align}$$ So the probability that no two friends are placed in adjacent seats is $$\frac{195}{\binom{12}{4}}=\frac{13}{33}$$ and the probability that at least two friends are placed in adjacent seats is $$1 - \frac{13}{33} = \frac{20}{33}$$

Answer 3

Alternative approach : Inclusion-Exclusion

See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

The complementary probability of no adjacent friends will be computed as

$$\frac{N}{D} ~: ~D = 12 \times 11 \times 10 \times 9.$$

Let $~S~$ denote the collection of possible distributions, without any regard to whether friends are together.

There are $~\displaystyle \binom{4}{2} = 6~$ possible pairing violations. Let:

• $~P_1~$ denote the pairing [A,B].
• $~P_2~$ denote the pairing [A,C].
• $~P_3~$ denote the pairing [A,D].
• $~P_4~$ denote the pairing [B,C].
• $~P_5~$ denote the pairing [B,D].
• $~P_6~$ denote the pairing [C,D].

For $~k \in \{1,2,\cdots,6\},~$ let $~S_k~$ denote the subset of $~S~$ where pair $~P_k~$ are adjacent. For example, $~S_1~$ denotes the subset where [A,B] are together, and there may or may not be other adjacent friends. Then the desired computation of $~N~$ is

$$N = |S| - |S_1 \cup \cdots \cup S_6|. \tag1 $$

Let $~T_0~$ denote $~|S|.$

Let $~T_1~$ denote $|S_1| + |S_2| + \cdots + |S_6|.$

For $~r \in \{2,3,4,5,6\},~$ let $~T_r~$ denote

$\displaystyle \sum_{1 \leq i_1 < i_2 < \cdots < i_r \leq 6} |S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_r}|.$

That is, $~T_r~$ denotes the sum of $~\displaystyle \binom{6}{r}~$ terms.

Then, in accordance with Inclusion-Exclusion theory, the computation in (1) above is equivalent to the following:

$$N = \sum_{r=0}^6 (-1)^{r+1}T_r. \tag2 $$

I am going to partition the seats into Section-1, Section-2, Section-3, Section-4, with these sections assigned the seats 1-3, 4-6, 7-9, 10-12, respectively.

The middle seats in these sections will be seats 2,5,8,11, respectively.

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$\underline{\text{Computation of} ~T_0}$

$T_0 = D = 12 \times 11 \times 10 \times 9 = 11880.$

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$\underline{\text{Computation of} ~T_1}$

To compute $~|S_1|,~$ note that [A,B] must be in the same section. So, since there are $~4~$ sections, you can reserve the factor of $~4,~$ and then assume, without loss of generality that $~[A,B]~$ are in section-1.

Within section-1, there are $~2~$ choices for which of A,B will be in seat 2. Once seat 2 is assigned, there are then $~2~$ choices for which of seats 1 or 3 will also be used by [A,B].

So, just focusing on [A,B], there are $~4 \times 2 \times 2 = 16~$ choices, for $~S_1.~$ Then, to complete the enumeration of $~S_1,~$ note that you then have $~10~$ choices for C and then $~9~$ choices for D.

So, $~|S_1| = 16 \times 90 = 1440.$

By considerations of symmetry,

$~|S_1| = |S_2| = \cdots = |S_6|.$

Therefore,

$$T_1 = 6 \times 1440 = 8640.$$

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$\underline{\text{Computation of} ~T_2}$

The $~\displaystyle \binom{6}{2} = 15~$ terms in the computation of $~T_2~$ will fall into two categories:

• Category-1
  The two pairings have a person in common.
  There are $~4~$ choices for which person is in both of the subsets, and then $~3~$ choices for which person is absent from both of the subsets.
  
  Therefore, $~12~$ of the $~15~$ terms are Category-1.
  Category-1 may be represented by $~|S_1 \cap S_2|,~$ which intersects [A,B] with [A,C].
  
  There are then $~4~$ choices for which section is utilized by A,B,C. Once this is established, there are two choices for assigning the outer seats in the section to B and C. Then, there will be $~9~$ choices for D.
  
  Therefore, the partial sum, represented by Category-1 is
  $12 \times 4 \times 2 \times 9 = 864.$

• Category-2
  The two pairings do not have a person in common.
  Then, A can be paired with either B,C, or D, with the remaining two also paired up. So, you can reserve a factor of $~3,~$ and then assume, without loss of generality that you are computing $~|S_1 \cap S_6|~$ which has [A,B] paired and [C,D] paired.
  
  From the previous section, it has already been established that [A,B] may be paired up in $~4 \times 2 \times 2 = 16~$ ways. There will then be $~3~$ sections left for [C,D] to be paired up, which can then be done in $~3 \times 2 \times 2 = 12~$ ways.
  
  Therefore, $~|S_1 \cap S_6| = 16 \times 12.$
  Therefore, the enumeration for Category-2 is
  $3 \times 16 \times 12 = 576.$

Therefore,

$$T_2 = 864 + 576 = 1440.$$

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$\underline{\text{Computation of} ~T_r ~: ~r \geq 3}$

First, consider the computation of $~T_3.~$ In this consideration, focus on attempting to extend one of the terms in $~T_2.~$

First, focus on attempting to extend a Category-1 intersection from the previous section. $~S_1 \cap S_2~$ is representative. So, you are starting with [A,B] and [A,C] both adjacent, which implies that A,B,C must completely fill a section of three seats.

If you try to extend it to $~S_1 \cap S_2 \cap S_4,~$ which represents the pairings [A,B], [A,C], [B,C] you see that this is impossible. That is, only one of A,B,C can have the middle seat of the pertinent section.

The only other way of extending $~S_1 \cap S_2~$ would be to also involve person D. This is also impossible, since you can not squeeze four people into a section that only has three seats.

Consideration of Category-2 extensions are similar. For example, consider starting with $~S_1 \cap S_6,~$ which represents the [A,B] and [C,D] pairings. Any attempt to extend this will involve one of A,B paired with one of C,D. This (again) means that all four people must be squeezed into one section of three seats. Again, this is impossible.

Therefore,

$$T_3 = 0.$$

Further, since the intersection of any 3 of $~S_1, S_2, \cdots, S_6,~$ must be the empty set, the intersection of more than 3 of $~S_1, S_2, \cdots, S_6,~$ must also be the empty set.

Therefore,

$$0 = T_4 = T_5 = T_6.$$

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$\underline{\text{Final Computation}}$

$$N = T_0 - T_1 + T_2 = 11880 - 8640 + 1440 = 4680.$$

$$D = 11880.$$

The complementary probability is

$$\frac{4680}{11880} = \frac{13}{33}.$$

Therefore, the probability of at least one occurrence of adjacent friends is

$$1 - \frac{13}{33} = \frac{20}{33}.$$