Question:hard

20 guests have to sit half on each side of a long table. Three particular guests desire to sit on a particular side, five others on the other side. In how many ways can the seating arrangement be made?

Show Hint

Split the 12 unrestricted guests between the two sides first, then arrange each side of 10.
Updated On: Jul 21, 2026
  • \( 10! \times \dfrac{10!}{7!} \times \dfrac{10!}{5!} \)
  • \( 12! \times \dfrac{10!}{7!} \times \dfrac{10!}{5!} \)
  • \( 20! \times 13! \times 15! \)
  • \( 20! \times \dfrac{7!}{3!} \)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Place each fixed group using direct permutations.
The 3 guests wanting a particular side can be placed among that side's 10 seats in $P(10,3) = \dfrac{10!}{7!}$ ways.
The 5 guests wanting the other side can be placed among that side's 10 seats in $P(10,5) = \dfrac{10!}{5!}$ ways.

Step 2: Fill the remaining 12 seats with the remaining 12 guests.
After placing the fixed guests, 7 seats remain on the first side and 5 seats remain on the second side, that is 12 seats in total, to be filled by the 12 leftover guests.
These 12 guests can be arranged into these 12 remaining seats in $12!$ ways.

Step 3: Multiply the three counts.
Total $= \dfrac{10!}{7!} \times \dfrac{10!}{5!} \times 12!$, the same expression reached by the combination method.

Final Answer:
Total ways $= 12! \times \dfrac{10!}{7!} \times \dfrac{10!}{5!}$, matching option (b). \[ \boxed{12! \times \dfrac{10!}{7!} \times \dfrac{10!}{5!}} \]
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