Arranging people in a jeep.

Question

There are $6$ seats in a Jeep, $3$ in the front and $3$ in the back. In how many ways $6$ persons of different heights can be seated so that everyone in the front is shorter than the person directly behind them?

What I tried is:

$(\binom{6}{2} × \binom{3}{1})$ choosing any $2$ people out of $6$ people. As they both will be of different height and one of them will be shorter than other. Now choose any one pair out of $3$ pairs of seats. In one pair there is one seat at front and one seat at back row.

$(\binom{4}{2} × \binom{2}{1})$ choosing any $2$ people out of $4$ remaining people. As they both will be of different height and one of them will be shorter than other. Now choose any one pair out of $2$ remaining pairs of seats. In one pair there is one seat at front and one seat at back row.

$(\binom{2}{2} × \binom{1}{1})$ choosing any $2$ people out of $2$ remaining people. As they both will be of different height and one of them will be shorter than other. Now choose any one pair out of $1$ pair of seats. In one pair there is one seat at front and one seat at back row.

so the final answer is $(\binom{6}{2} × \binom{3}{1}) × (\binom{4}{2} × \binom{2}{1}) × (\binom{2}{2} × \binom{1}{1}) = 540$.

But the answer given on many online sites is $90$.

Where am I making mistake?

Answer 1

Lots of good things in this near-solution! There's only one mistake: each legal seating configuration actually arises $6=3!$ different ways in the construction, depending on the order in which the pairs of seats are chosen. When that factor of $6$ is accounted for, the answer of $540/6 = 90$ is right on the money.

Here's another solution: group all possible seating configurations into sets of $8$, where two configurations are in the same set precisely when each pair of seats contains the same pair of people (possibly switched with each other). In each set of $8$, there is precisely $1$ where the shorter person in each pair is in the front seat of the pair of seats. So the total number of valid configurations is the same as the number of sets of $8$, which is $6!/8 = 720/8 = 90$.

Answer 2

As indicated in Greg Martin's answer, you are currently overcounting because every configuration is considered $6$ times.

Imagine starting with the pair $(A, B)$ and assigning them to the left-most chairs. Then, you choose the pair $(C, D)$ and assign them to the right-most chairs. You would be able to arrive at the same configuration if you had started with the pair $(C, D)$ and assigned them to the right-most chairs, only then to choose the pair $(A, B)$ and assign them to the left-most chairs.

To arrive at the correct answer, I would use the following reasoning:

• Select a pair of people to assign to the left-most seats in ${6 \choose 2} = 15$ ways
• Select a pair of people to assign to the middle seats in ${4 \choose 2} = 6$ ways
• Select a pair of people to assign to the right-most seats in ${2 \choose 2} = 1$ ways

Thus, the number of valid configurations equals:

$${6 \choose 2} {4 \choose 2} {2 \choose 2} = 15 \cdot 6 \cdot 1 = 90$$

Answer 3

Alternative approach:

Use 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.

Label the people in ascending order, by height, as $~P_1, ~P_2, ~\cdots, ~P_6.~$ This implies that $~P_i~$ is shorter than $~P_j~$ if and only if $~i < j.$

Following the syntax in the 2nd Inclusion-Exclusion link,
let $~S~$ denote the set of all possible ways of seating the $~6~$ people.

Let

• $S_1~$ denote the subset of $~S,~$ where the person in the front row, left most seat is taller than the person in the back row, left most seat.

• $S_2~$ denote the subset of $~S,~$ where the person in the front row, middle seat is taller than the person in the back row, middle seat.

• $S_3~$ denote the subset of $~S,~$ where the person in the front row, right most seat is taller than the person in the back row, right most seat.

Then, the desired computation is

$$|~S~| - | ~S_1 \cup S_2 \cup S_3 ~|. \tag1 $$

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Let

• $T_0~$ denote $~| ~S ~|.$

• $T_1~$ denote $~| ~S_1 ~| + | ~S_2 ~| + | ~S_3 ~|.$

• $T_2~$ denote $~| ~S_1 \cap S_2 ~| + | ~S_1 \cap S_3 ~| + | ~S_2 \cap S_3 ~|.$

• $T_3~$ denote $~| ~S_1 \cap S_2 \cap S_3 ~|.$

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

$$[~T_0 + T_2 ~] - [~T_1 + T_3 ~]. \tag2 $$

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$$T_0 = | ~S ~| = 6! = 720.$$

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To compute $~| ~S_1 ~|,~$ consider that there are $~\displaystyle \binom{6}{2} = 15~$ ways for selecting the two people that will sit in the left most seats. Once this is determined, there is only one way of ordering those two people so that the taller person is in front.

Therefore, $~\displaystyle | ~S_1 ~| = 15 \times 4! = 360.$

By considerations of symmetry, $~| ~S_1 ~| = | ~S_2 ~| = | ~S_3 ~|.$

Therefore,

$$T_1 = 3 \times 360 = 1080. \tag2 $$

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The analysis in this section will be very similar to the analysis in the previous section.

To compute $~| ~S_1 \cap ~S_2 ~|,~$ consider that there are $~\displaystyle \binom{6}{2} \times \binom{4}{2} = 90~$ ways for selecting the two people that will sit in the left most seats, and then selecting the two people that will sit in the middle seats. Once this is determined, there is only one way of ordering those four people so that in both the left most seats and the middle seats, the taller person is in front.

Therefore, $~\displaystyle | ~S_1 \cap ~S_2 ~| = 90 \times 2! = 180.$

By considerations of symmetry, $~| ~S_1 \cap S_2 ~| = | ~S_1 \cap S_3 ~| = | ~S_2 \cap S_3 ~|.$

Therefore,

$$T_2 = 3 \times 180 = 540.$$

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Similar to the analysis in the previous section, to compute $~T_3 = | ~S_1 \cap S_2 \cap S_3 ~|,~$ consider that there are

$$T_3 = \binom{6}{2} \times \binom{4}{2} \times \binom{2}{2} = 90$$

ways of selecting and seating the people.

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Therefore, the final computation is

$$[~T_0 + T_2 ~] - [~T_1 + T_3 ~] $$

$$= [~720 + 540 ~] - [ ~1080 + 90~] = 90.$$