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Permutations and combinations: RVHS 2023 Promo Q2

River Valley High School2023 PromoPromo Q2●●○ Standard6 marks

What this question tests

Permutations and combinations.

Question

At a sit-down dinner, 2 adults, 4 boys and 4 girls are to be seated at a round table with ten identical chairs. Find the number of different ways that they can be seated if

  1. there are no restrictions; [1]

  2. not all the 4 boys are seated together; [2]

  3. the 2 adults are seated together, and a girl is on each side of them. [3]

Show full worked solution▾

(i) For a round table with 10 identical chairs and no restrictions: No. of ways=(10−1)!=362 880.\begin{aligned} \text{No. of ways} &= (10 - 1)!\\ &= 362\,880. \end{aligned}

(ii) No. of ways with not all 4 boys together == (no. of ways without restriction) −- (no. of ways with all 4 boys together). No. of ways=362 880−(7−1)!×4!=345 600.\begin{aligned} \text{No. of ways} &= 362\,880 - (7 - 1)!\times 4!\\ &= 345\,600. \end{aligned}

(iii) Treat the 2 adults with a girl on each side as a single unit GAAGGAAG.

Choose 2 girls out of 4 to sit on each side and arrange them with the adults: (42)×2!×2!\binom{4}{2}\times 2! \times 2!.

Treat GAAGGAAG as one super-unit and arrange with the other 6 people around the table: (7−1)!(7 - 1)!. No. of ways=(7−1)!×(42)×2!×2!=17 280.\begin{aligned} \text{No. of ways} &= (7 - 1)! \times \binom{4}{2}\times 2! \times 2!\\ &= 17\,280. \end{aligned}

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