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Eva8 [605]
3 years ago
8

In how many 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?

Mathematics
1 answer:
balandron [24]3 years ago
7 0
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
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