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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!(n−1)!=3!3!(3−3)!=6(n, r)=\frac{n !}{R !(n-1) !}=\frac{3 !}{3 !(3-3) !}=6 different ways.

Studdy Solution

STEP 1

1. The books are considered distinct within their respective categories (English, Chemistry, Math).
2. When books of the same subject must be kept together, they can be treated as a single item for the purpose of arranging, but their internal arrangement also matters.
3. The factorial notation n!n! represents the product of all positive integers from 11 to nn.

STEP 2

1. Calculate the number of arrangements with no restrictions.
2. Calculate the number of arrangements where books of each subject must be kept together.

STEP 3

Calculate the number of different arrangements for the 9 books with no restrictions using the formula for permutations of distinct objects.
n!=9! n! = 9!

STEP 4

Compute the value of 9!9!.
9!=9×8×7×6×5×4×3×2×1 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1

STEP 5

Simplify the computation to find the total number of arrangements with no restrictions.
9!=362,880 9! = 362,880

STEP 6

Consider each set of books of the same subject as a single item and calculate the number of arrangements for these items.
There are 3 subjects, so treat each subject as a single item, leading to 3!3! arrangements.

STEP 7

Compute the value of 3!3! for the arrangements of the subjects.
3!=3×2×1 3! = 3 \times 2 \times 1

STEP 8

Simplify the computation to find the number of arrangements for the subjects.
3!=6 3! = 6

STEP 9

Calculate the number of different arrangements within each subject group.
For English books: 2!2! arrangements. For Chemistry books: 4!4! arrangements. For Math books: 3!3! arrangements.

STEP 10

Compute the values of 2!2!, 4!4!, and 3!3!.
2!=2×1 2! = 2 \times 1 4!=4×3×2×1 4! = 4 \times 3 \times 2 \times 1 3!=3×2×1 3! = 3 \times 2 \times 1

STEP 11

Simplify the computations to find the number of arrangements within each subject group.
2!=2 2! = 2 4!=24 4! = 24 3!=6 3! = 6

STEP 12

Multiply the number of arrangements for each subject group by the number of arrangements of the subjects to get the total number of different arrangements when the books of each subject must be kept together.
Total arrangements=3!×2!×4!×3! \text{Total arrangements} = 3! \times 2! \times 4! \times 3!

STEP 13

Substitute the computed values into the total arrangements formula and simplify.
Total arrangements=6×2×24×6 \text{Total arrangements} = 6 \times 2 \times 24 \times 6

STEP 14

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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