Number of ways $6$ persons of different height be seated So that every one in front is shorter? In a jeep there are $3$ seats in front & $3$ at back.
Number of ways $6$ persons of different height
be seated So that every one in front is shorter
than the person directly behind him
MY ATTEMPT:
Suppose the jeep is like this with the upper row as the front row.

The tallest three among the 6 can be told to enter the back row.Morever they can be arranged in $3!$ ways.Similarly the remaining 3 people can be told to occupy the front row.They too may be arranged in $3!$ ways.So answer should be $3!*3!=6*6=36$
However this answer is incorrect as per the answer key in my book.Where did I go wrong?
 A: You are incorrectly assuming that the tallest $3$ must sit in the back and the shortest $3$ must sit in the front.
Let's give these $6$ people values: $1, 2, ..., 6$ denoting their height (say $6$ is the tallest). The question can be reformulated as asking for the number of ways we can make $3$ pairs of people (and then seat these $3$ pairs into $3$ columns). We note that when we select the $3$ pairs, each pair has exactly one way of sitting in a selected column (the larger number sits in the back, the smaller number sits in the front).
As a counterexample to your assumption, consider the pairings $(6,5)$, $(4,3)$, and $(2,1)$, where $6$, $4$, and $2$ sit in the back row, and $5$, $3$, and $1$ sit in the front row.
Now, let's solve the problem: There are ${{6}\choose{2}}=15$ ways to choose the first pair, and ${{4}\choose{2}}=6$ ways to choose the second pair. Therefore, there are $15\times6=90$ ways to choose the $3$ pairs. We've already ordered the pairs (the first pair chosen sits in the first column, the second pair sits in the second, etc), so our final answer is $90$.
A: Let $[6]$ be the set of persons, named by their heights. Then $1$ can choose any of the three front seats and any of five persons to sit directly behind him. The smallest of the remaining four persons can choose either of the two left over front seats and any of three remaining persons to sit directly behind him. The last two persons then have no choice anymore. Thus we obtain a total of $3\cdot 5\cdot 2\cdot 3=90$ possibilities.
A: You can choose two people to sit on the driver's side in ${6\choose2}=15$ ways, then choose two people to sit on the passenger's side in ${4\choose2}=6$ ways, with the remaining two people to sit in the middle. Once you've made these choices, the taller person in each pair must sit in the back, so there is a total of $15\cdot6=90$ seating assignments in all.
A: The shortest person can sit in any of the three front seats. The second shortest person can sit in the two unoccupied front seats or behind the shortest person . The third person can sit behind either of the other two or in the unoccupied seat in the front unless two sat behind one, in which case, he only has two options. If the three shortest people sat in the front, then the three tallest people can sit in any order in the back in six ways. If one of the shortest three sat behind the other, then the fourth shortest can sit in the open front seat or behind the shorter three that does not have anyone sitting behind him. Five can sit in either of the back two spots if the front is filled up, and he has to fill up the front if there is an open spot. Finally, six has to sit in the remaining spot. You should be able to make something from this.
A: In your 36 arranges, you have not considered the next one:
135
246
A way to proceed is the following:
It is easy to see that the number of words you can do by using the letters aabbcc is $\frac{6!}{2!2!2!}=90$.
Now, let see that this is the same answer to your problem.
Put the six people in a row in order 123456 and choose one of the words described above, for example, abaccb.
Now, put it below the row
\begin{align*}
123456\\
abaccb
\end{align*}
and seat on the first row of the jeep those with a, seat on the second row the b's, and the third row by the c's.
124
365.
This is a 1-1 correspondence. Then, the answer is 90.
