Three married couples should be arranged on 6 different seats in one row of the cinema hall with 12 numbered seats Three married couples should be arranged on 6 different seats in one row of the cinema hall with 12 numbered seats. The number of ways this can be done so that each pair is arranged on two adjacent seats is equal to ?
Until now i only figured out that each couple has to be arranged among themselves in 2! ways and each couple can be seated next to each other in 3! ways.
I can't proceed further, I just can't understand the logic behind ?
 A: Seat the couples in $3$ "double seaters" , call them $D$
Each couple can be arranged in $2^3$ = $8$ ways on the  $D's$
Take a $D$ and place it in the $7$ gaps including ends between the chairs.
$\uparrow\square\uparrow\square\uparrow\square\uparrow\square\uparrow\square\uparrow\square\uparrow$
Each time a $D$ is placed, one more gap is created for the next,
so # of ways $ = 2^3\cdot7\cdot8\cdot9 = 4032$
A: In addition to what you have done, we have to determine which seats are occupied.
Method 1:  Let $x_1$ be the number of seats to the left of the first couple, let $x_2$ be the number of seats between the first and second couples, let $x_3$ be the number of seats between the second and third couples, and let $x_4$ be the number of seats to the right of the fourth couple.  Since the three couples will occupy six of the twelve seats, there are $12 - 6 = 6$ unoccupied seats.  Hence,
$$x_1 + x_2 + x_3 + x_4 = 6 \tag{1}$$
is an equation in the nonnegative integers.  A a particular solution of equation $1$ corresponds to the placement of $4 - 1 = 3$ addition signs in a row of $6$ ones.  For instance,
$$1 1 + + 1 + 1 1 1 1$$
corresponds to the solution $x_1 = 2, x_2 = 0, x_3 = 1, x_4 = 4$ (the leftmost two seats are unoccupied, the first two couples sit next to each other in the next four seats, there is another unoccupied seat to the right of the second couple, the third couple is seated, and the last four seats are unoccupied).  The number of solutions of equation $1$ is the number of ways we can place $4 - 1 = 3$ addition signs in a row of $6$ ones, which is
$$\binom{6 + 4 - 1}{4 - 1} = \binom{9}{3}$$
since we must choose which three of the nine positions required for six ones and three addition signs will be filled with addition signs.
Since there are $\binom{9}{3}$ ways to select which seats will be unoccupied (and, thus, which seats will be occupied), $3!$ ways of arranging the three couples in the pairs of adjacent seats which are to be occupied, and $2!$ ways to arrange each couple in its pair of seats, there are
$$\binom{9}{3}3!2!2!2!$$
possible seating arrangements for the couples.
Method 2:  @angryavian made a nice suggestion in the comments.  Suppose we have six blue and six green chairs.  Take two of the green chairs and weld them together.  Repeat until there are three pairs of welded green chairs.  Now we have nine objects to arrange, the six blue chairs and three pairs of welded green chairs.  Choose three of the nine positions for the pairs of green chairs, which can be done in $\binom{9}{3}$ ways.  Number the chairs from left to right.  The first couple will sit in the first pair of adjacent green chairs, the second couple will sit in the next two adjacent green chairs, and the third couple will sit in the last pair of adjacent green chairs.  There are $3!$ ways of arranging the couples and $2!$ ways of arranging each couple within their pair of green chairs.  We again obtain
$$\binom{9}{3}3!2!2!2!$$
possible seating arrangements.
