number of ways without restriction:
9!/9 = 8!
now let's put Sam between Ron and Ed
RSE or ESR, so that's 2 ways
let's consider RSE as one grouping, that means we still have 6 others to arrange
number of ways = 2x7!/7
number of ways for the stated condition
= 8! - 2(7!)/7
= 8! - 2(6!)
= 8x7x6! - 2x6!
= 6!(56-2)
= 54 x 6!
= 38880
How many ways can 9 people be arranged to sit at a round table if Sam can not sit between Ron and Ed?
My solution: 7!
Thanks for your help.
1 answer