Number of ways in which 5 boys and 5 girls can be seated alternatively on a round table if a particular boys and a particular girl are never adjacent to each other in any arrangement, is

Question

Number of ways in which 5 boys and 5 girls can be seated alternatively on a round table if a particular boys and a particular girl are never adjacent to each other in any arrangement, is (A) 288 (B) 552 (C) 1584 (D) 172

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/cadb03711611304a6b3efa175dc19ff8-.png" /> B1 mid G1 always separated. 5 boys can be seated in 4 ! now for G1 can not sit between B 1 and B 2 or B 1 and B 5 and therefore can set only in 3 ways and the remaining 4 girls in 4 ! ways. Total ways (4 !)(3)(4 !)=576 × 3=1728

Q. Number of ways in which 5 boys and 5 girls can be seated alternatively on a round table if a particular boys and a particular girl are never adjacent to each other in any arrangement, is

288

552

1584

172

$B_{1} ∣ G_{1}$ always separated.
5 boys can be seated in 4 ! now for $G_{1}$ can not sit between $B_{1}$ and $B_{2}$ or $B_{1}$ and $B_{5}$ and therefore can set only in 3 ways and the remaining 4 girls in 4 ! ways.
Total ways $(4!) (3) (4!) = 576 \times 3 = 1728$

Q. Number of ways in which 5 boys and 5 girls can be seated alternatively on a round table if a particular boys and a particular girl are never adjacent to each other in any arrangement, is

288

552

1584

172

B1​∣G1​ always separated. 5 boys can be seated in 4 ! now for G1​ can not sit between B1​ and B2​ or B1​ and B5​ and therefore can set only in 3 ways and the remaining 4 girls in 4 ! ways. Total ways (4!)(3)(4!)=576×3=1728

$B_{1} ∣ G_{1}$ always separated.
5 boys can be seated in 4 ! now for $G_{1}$ can not sit between $B_{1}$ and $B_{2}$ or $B_{1}$ and $B_{5}$ and therefore can set only in 3 ways and the remaining 4 girls in 4 ! ways.
Total ways $(4!) (3) (4!) = 576 \times 3 = 1728$