Eight guests have to be seated 4 on each side of a long rectangular ta...
Solution:
Given, 8 guests have to be seated 4 on each side of a long rectangular table.
Two particular guests desire to sit on one side of the table and 3 on the other side.
Let us consider the two particular guests as A and B, and the remaining guests as C, D, E, F, G, and H.
Now, A and B have to sit on one side of the table, and the remaining three guests have to sit on the other side.
The number of ways in which A and B can be seated on one side of the table is 2! = 2.
The number of ways in which the remaining three guests can be seated on the other side of the table is 3! = 6.
Now, we have to arrange the remaining 3 guests on the other side of the table.
The number of ways in which 3 guests can be seated on one side of a rectangular table is (3-1)! = 2.
Therefore, the total number of ways in which the seating arrangements can be made is 2 x 6 x 2 = 24.
Since there are two identical sides of the table, the total number of ways in which the seating arrangements can be made is 24/2 = 12.
Hence, the correct answer is option (d) 1278.
Note: It is important to break down the problem into smaller parts and use the concept of permutation and combination to arrive at the solution. Also, it is necessary to consider all the given conditions while solving the problem.
Eight guests have to be seated 4 on each side of a long rectangular ta...
3c1×4!×4!= 1728
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