1. 12 Friends are to sit in a circular counting of 12 seats,A and B are sworn en

Question

1. 12 Friends are to sit in a circular counting of 12 seats,A and B are sworn enemies and can never sit together while two of the C and D are sweethearts and would always want to sit together., how many seating arrangements are possible?

Hi, I made mine
Case 1 : No restriction 11!
Case 2: Sweet hearts only
(12-2(sweethearts)+1(count sweethearts as one) -1(circular arrangement))x2!
=13!x2!
Case 3: enemies (same as sweetheart)
13!x2!
-----> 14! -(13!x2)!

2. What if Friend A & B doesnt show up?

Explanation / Answer

1. Since the two sweethearts are always together, let us take them as one BIG person.

Thus there are 10 + 1 = 11 people (big or small) and they can be arranged in (11-1)! = 10! ways.

Futher the two sweethearts can sit among themselves in 2! ways.

Thus the 11 'people' can be seated in 10! * 2! ways.

(Note: The number of ways of seating n people around a circle is (n-1)!)

Let us take the additional case of the two enemies sitting together.

There are now 2 big people and 10 people in all.

These can be arranged in 9! ways and the two big people can be arranged in 2! ways each.

There are 9! * 2! * 2! ways to arrange them.

Since the two enemies should not be seated together, the number of ways this is possible

= 10! * 2! - 9! * 2! *2!

= 9! * 2! * (10 - 2!)

= 9! * 2! * 8

= 5806080.

2. Since A and B don't show up we are left with 10 people.

Of these C and D constitute a big person.

Therefore, there are 9 'people' and these can be seated around the table in (9 -1)! = 8! ways.

C and D can be seated among themselves in 2! ways.

Total number of ways to seat the friends = 8! * 2! = 80640.

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