Question

There are 7 men and 5 women in a park. The number of ways of arranging them around a circular path such that 4 particular persons which include 2 particular men and 2 particular women never stand together is:

• A. \(11879(8!)\)
• B. \(966(8!)\)
• C. \(986(8!)\)
• D. \(494(4!)(8!)\)

Hint:

Whenever a circular permutation contains a condition like “never together”, first calculate total arrangements and then subtract arrangements where all specified persons remain together.

Answer 1

The Correct Option is B

Concept: For circular arrangements of \(n\) distinct persons, \[ \text{Number of arrangements}=(n-1)!. \] When some persons are together, they are treated as one block.

Step 1: Calculate total circular arrangements.
Total persons \[ =7+5=12. \] Therefore, \[ \text{Total arrangements} =(12-1)! =11!. \]

Step 2: Count arrangements in which the four particular persons are together.
Treat the four particular persons as one block. Thus total units become \[ 8+1=9. \] Hence circular arrangements are \[ (9-1)! = 8!. \] The four persons inside the block can be arranged in \[ 4! \] ways. Therefore, \[ N(\text{all four together}) = 4!(8!). \]

Step 3: Apply complementary counting.
Required arrangements \[ = 11!-4!(8!). \] Now, \[ 11! = 11\times10\times9(8!) = 990(8!). \] Hence, \[ 990(8!)-24(8!) = 966(8!). \] \[ \boxed{966(8!)} \]