There are nine differently-titled books on a shelf. Four books have red covers, while the other five have black covers. How many different arrangements of the books can there be, if
(a) there are no restrictions on the order in which the books are placed?
Is it 9! ?
(b) all the black-covered books must be together?
Is it 5! ?
(c) all the black-covered books must be together and all the red-covered books must also be together?
is it 2! ?
(d) black- and red-covered books must alternate?
is it 4!5! ?
looks good
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Yep spot on
(a) When there are no restrictions on the order in which the books are placed, we can arrange the books freely. Since we have 9 books in total with different titles, the number of different arrangements can be calculated using the factorial function. Therefore, the answer is 9!.
(b) When all the black-covered books must be together, we can treat the group of black-covered books as a single entity. This reduces the problem to arranging the 5 groups: the group of black-covered books and the 4 individual red-covered books. Considering the group of black-covered books as one unit, we now have 5 units to arrange. The number of different arrangements can be calculated using the factorial function, so the answer is 5!.
(c) When all the black-covered books must be together, and all the red-covered books must also be together, we can treat both sets as single entities. Now, we have two groups, the black-covered books and the red-covered books, to arrange. So, the number of different arrangements is 2!.
(d) When black- and red-covered books must alternate, we can first arrange the black-covered books and then arrange the red-covered books in between them. Since there are 4 black-covered books and 5 red-covered books, we can arrange the black-covered books in 4! ways and the red-covered books in 5! ways. Multiplying these two possibilities, we get the total number of arrangements as 4! * 5!.