Question

In a board meeting, 8 delegates from number 1 to 8 are sitting around a circular table. Number 4 and 5 always sit together and number 1 always sits next to number 4. If number 8 always sits exactly opposite number 3, in how many ways can the seating be done?

• A. 6!
• B. ⁶C₂ x 3
• C. 4x3!
• D. 4! x 2

Answer 1

The correct option is C

4x3!

Let's fix number 8 at a slot at the table, automatically, number 3 also gets fixed. Out of the remaining 6 delegates, 4 and 5 will be together and can be arranged in 2! Ways amongst each other. Also for each of these ways, number 1 can sit next to 4 in only one way.

Now 1,4,5 can be placed only in 2 ways(See below).

This question now becomes the arrangement of the remaining 3 people $\times$ 4= 3! $\times$ 4. Option (c)