Permutations & Combinations
Constrained seating arrangement
MJAT_TS2_P2
Grade 12
Question:
Ten individuals — 2 Americans ($A_1,A_2$), 2 Russians ($R_1,R_2$), 2 Chinese ($C_1,C_2$), 2 Dubaians ($D_1,D_2$), and 2 Sentinelese Indians ($In_1,In_2$) — are seated in a row of 10 labeled chairs with constraints:
1. $A_1$ and $A_2$ occupy the two extreme seats.
2. Dubaians sit together as a block, placed directly between either both Russians ($R$-$D_1D_2$-$R$) or both Chinese ($C$-$D_1D_2$-$C$).
3. $A_1$ (Nationalist) refuses to sit next to any Russian.
The total number of valid seating arrangements is:
Step-by-Step Solution
Key Concept: Case 1: Dubaians between Russians (R-D₁D₂-R). Treat $R_1D_1D_2R_2$ as one block of 4. Remaining seats (positions 2-9, 8 seats): this block (4), $C_1,C_2,In_1,In_2$ fill 4 more. $A_1$ at one extreme cannot be next to $R$ — check which end $A_1$ is at and whether adjacent seat has a Russian.
From solution: Case(1) block adjacent: $2!\times 2!\times\frac{4!}{2!2!}\times 2! = 32\times(4!)$... Total across both cases and arrangements = $56\times 4! = 56\times 24 = 1344$. Answer: $\mathbf{1344}$.
Correct Answer: 1344