Question

12 persons are to be seated at a square table, three on each side. 2 persons wish to sit on the north side and two wish to sit on the east side. One other person insists on occupying the middle seat (which may be on any side). Find the number of ways in which they can be seated.

Answer 1

In this question, we are given the arrangements of 12 persons at a square table along with certain conditions. We have been asked to find out the number of ways in which the persons can be seated following the given conditions. Start with dividing the problem into 2 cases: 1 with north and east and the other with south and west. In both the cases, settle the person who has to sit in the middle first. Then, give choices to the persons who want to sit on a particular side. After this, seat the remaining people.

Complete step-by-step answer:
We have to find out the number of ways in which 12 persons can be seated following the given conditions. We will have to find the answer as per the person sitting in the middle.
A man wants to sit on the middle seat of any side. There will be 2 cases as follows:
Case 1:
If the person sits on the North or East side.
We already have certain conditions for the east and north side. So, if the man sits in the middle of these sides, he will have to comply with the other conditions as well.
If a person sits on the middle seat of either north or east, then two of them will have to sit on either of his sides in $^2{P_2}$ ways. Then the other two persons can select their seats in $^3{P_2}$ ways. This can be done in two ways as the person can sit on the north or east side. $\Rightarrow 2{ \times ^3}{P_2}{ \times ^2}{P_2}$
After this, the remaining 7 persons can sit anywhere in 7 places in $7!$ ways.
Therefore, total ways = $2{ \times ^3}{P_2}{ \times ^2}{P_2} \times 7!$
$\Rightarrow 2 \times 6 \times 2 \times 7!$
$\Rightarrow 24 \times 7!$ ways
Case 2:
If the person sits on the South or West side.
In this case, the people who want to sit on the north and east side will have 3 options each which they can choose in $^3{P_2}$ ways each. On the other hand, now the person who wants to sit in the middle has 2 options- south side or west side. The remaining 7 persons can sit in $7!$ ways.
Therefore, total ways = $2{ \times ^3}{P_2}{ \times ^3}{P_2} \times 7!$
$\Rightarrow 2 \times 6 \times 6 \times 7!$
$\Rightarrow 72 \times 7!$ ways
Hence, total ways = $24 \times 7! + 72 \times 7!$
$\Rightarrow 7!(24 + 72)$
$\Rightarrow 96 \times 7!$
$\Rightarrow 483,840$ ways

There are a total of 483840 ways in which they can be seated.

Note: We observe that in mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. The word “permutation” also refers to the act or process of changing the linear order of an ordered set. Permutations differ from combinations, which are selections of some members of a set regardless of order. It is necessary to divide the question into two cases as in both the cases, the calculations and approach will be different and cannot be calculated together.