The 5 people can seat in a row in 5! ways But we need to exclude the ways <span>that ann and bob are seated next to each other which is = 4! * 2! </span> So, the number of <span>ways can ann, bob, chuck, don and ed be seated in a row such that ann and bob are not seated next to each other = 5! - 4! * 2! = 72 </span> <span>===================================================== </span> <span>Another solution: </span> <span>If ann seated in one of the ends, the number of ways = 3*2 </span> <span>If ann didn't seat in one of the ends , the number of ways = 2*3 </span> So, the total number of <span>ways that can <span>ann, bob be seated = 3*2 + 2*3 = 12 </span></span> The remaining persons can seat with a number of ways = 3! = 6 So, the total ways that the five persons can seat = 12*6 = 72
