Question 1141051: Tell me if I'm correct or wrong. If wrong, kindly give me an explanation, thank you.
In how many ways can I arrange 3 different mathbooks and 5 different history books on my bookshelf, if I require there to be math on both ends?
My answer would be 3*5!*2! = 720
on both ends we could choose 3 math books and the preceding has 5 arrangement of
history books. The remaining would be followed by 2 arrangement of math books..
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website!
I think you are incorrect.
There are 3! = 6 distinct ways to have math books on the ends [M1...M2, M1...M3, M2...M3, M2...M1, M3...M1, and M3...M2]. This assumes that M1...M2 is distinct from M2...M1, which I think is reasonable. Once the two end books are chosen, the 3rd math book is mixed in with the history books. There are 6! distinct ways to arrange those six books (or you could say 5! ways to arrange the history books and then multiply by 6 for the different "spaces" the 3rd math book can occupy around those 5 books: 5!*6 = 6! ).
3!*6! = 6*720 = 4320 ways in all.
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