(Circular Permutations) In how many ways can 7 people { A, B, C, D, E, F, G } be

Question

(Circular Permutations) In how many ways can 7 people { A, B, C, D, E, F, G } be seated at a round table if

(a) A and B must not sit next to each other;

(b) C, D, and E must sit together (i.e., no other person can sit between any of these three)?

(c) A and B must sit together, but neither can be seated next to C or D.

Consider each of these separately.

Hint: Conceptually, think of the groups of two or three people as one "multi-person" entity in the overall circular arrangement. It may help to draw a diagram, fixing a particular person at the top of the circle (thereby eliminating the duplicates due to rotations).

Explanation / Answer

(2) Consider C, D and E as one individual unit, now we have 4 people left, so total we have 4 + 1 ( C, D and E) = 5 units but in circular permutation, its (n - 1)! = which is 4!. And C, D and E can exchange their place with each other.

So answer would be = 4! × 3! = 24 × 6 = 144

(3) consider AB as a single unit.

Now we will find in how many ways AB can sit next to C or D. So their are 2 possible ways so do so.

1. (AB) C D E F G = 6 units, but as its circular permutation its = 5!, and A and B can exchange their places, so its 2 × 5! = 240

2. E F G C D (AB) = 6 units, but as its circular permutation its = 5! And A and B can exchange their places, so its 2 × 5! = 240

Total ways = 240 + 240 = 480

So total possible ways of A and B together but not next to C or D = 720 - 480 = 240

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