Math  /  Arithmetic

Question Arrange 9 books (2 English, 4 Chemistry, 3 Math) on a shelf. (a) No restrictions: n=a9p9!362,880n=a_{9} p_{9} ! \equiv 362,880. (b) Books on each subject must be kept together: (n,r)=n!R!(n1)!=3!3!(33)!=6(n, r)=\frac{n !}{R !(n-1) !}=\frac{3 !}{3 !(3-3) !}=6 different ways.

Studdy Solution
Calculate the final result for the total number of different arrangements when the books of each subject must be kept together.
Total arrangements=6×2×24×6=1728 \text{Total arrangements} = 6 \times 2 \times 24 \times 6 = 1728
The solution to the problem is: a. With no restrictions, there are 362,880362,880 different arrangements. b. With the restriction that books of each subject must be kept together, there are 17281728 different arrangements.

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