4. Eight guests have to be seated 4 on each side of a long rectangular...
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.
To find: The number of ways in which the sitting arrangements can be made.
We can divide this problem into two cases:
Case 1: Two particular guests sit on the same side of the table.
In this case, we have to select 2 guests out of 8 to sit on one side of the table, which can be done in 8C2 ways. The remaining 6 guests can be seated on the other side of the table in 6! ways. Also, we can arrange the two groups of guests in 2! ways.
Therefore, the total number of ways in which the sitting arrangements can be made in this case is:
8C2 × 6! × 2! = 201,600
Case 2: Two particular guests sit on opposite sides of the table.
In this case, we have to select 2 guests out of 8 to sit on one side of the table, which can be done in 8C2 ways. The remaining 6 guests can be seated on the other side of the table in 6! ways. Also, we can arrange the two groups of guests in 2! ways.
Out of the 3 remaining guests, we have to select 2 to sit on the same side of the table as the 2 particular guests. This can be done in 3C2 ways. The selected guests can be arranged in 2! ways.
Therefore, the total number of ways in which the sitting arrangements can be made in this case is:
8C2 × 6! × 2! × 3C2 × 2! = 483,840
Total number of ways = Case 1 + Case 2 = 201,600 + 483,840 = 685,440
Hence, the correct option is (d) 1278.
4. Eight guests have to be seated 4 on each side of a long rectangular...
3C1 × 4! × 4! = 1728
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