Eighteen guests have to be seated, half...

MathematicsCombinations and SelectionJEE Advanced 1991Easy

Eighteen guests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other side. Determine the number of ways in which the sitting arrangements can be made.

Answer:

^{11} C_{5} \times 9! \times 9!

Visualized Solution (Hindi)

Visualizing the Table

Total Guests: $18$
Seats per side: $\frac{18}{2} = 9$
Let the sides be Side A and Side B.

Identifying Constraints

Guests fixed for Side A: $4$
Guests fixed for Side B: $3$
Remaining Guests: $18 - (4 + 3) = 11$

Calculating Remaining Slots

Available slots on Side A: $9 - 4 = 5$
Available slots on Side B: $9 - 3 = 6$
We need to select $5$ guests out of $11$ for Side A.

Selection Process: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }^{11} C_{5}

Ways to select $5$ guests for Side A: $^{11} C_{5}$
Remaining $6$ guests are automatically assigned to Side B in $^{6} C_{6} = 1$ way.

Arrangement on Each Side

Arrangement of $9$ guests on Side A: $9!$
Arrangement of $9$ guests on Side B: $9!$
Total arrangements for both sides: $9! \times 9!$

Final Total Arrangement

Total Ways = Selection $\times$ Arrangement
Total Ways = $^{11} C_{5} \times 9! \times 9!$
Final Result: $^{11} C_{5} \times (9!)^{2}$

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