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Four Mathematics books, three History books, three Science books and two Biology books are arranged randomly on a shelf. What is the probability that: 23.1 all books of the same subject land up next to each other? 23.2 just the History books will be together?

Mathematics

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2 parts
23.1

111,550\frac{1}{11,550}

23.2

122\frac{1}{22}

Step-by-step solution

10 steps

Right Nqøbizithå🤍🩺, let's solve this probability problem about arranging books.

First, let's determine the total number of books and the total number of ways to arrange them.

  • Mathematics books: 4
  • History books: 3
  • Science books: 3
  • Biology books: 2 Total number of books = 4+3+3+2=124 + 3 + 3 + 2 = 12 books.

Assuming all books are distinct (e.g., different titles even if of the same subject), the total number of ways to arrange 12 books on a shelf is 12!12!.

N=12!=479,001,600N = 12! = 479,001,600

Part 23.1· 5 steps

Determine the probability that all books of the same subject land up next to each other.

  1. Treat each subject group as a single block.

    We have 4 blocks: (4 Math books), (3 History books), (3 Science books), (2 Biology books).

  2. Arrange these 4 blocks.

    The number of ways to arrange these 4 distinct blocks is 4!4!.

    4!=4×3×2×1=244! = 4 \times 3 \times 2 \times 1 = 24

  3. Arrange the books within each block.

    • The 4 Mathematics books can be arranged in 4!4! ways (4×3×2×1=244 \times 3 \times 2 \times 1 = 24).
    • The 3 History books can be arranged in 3!3! ways (3×2×1=63 \times 2 \times 1 = 6).
    • The 3 Science books can be arranged in 3!3! ways (3×2×1=63 \times 2 \times 1 = 6).
    • The 2 Biology books can be arranged in 2!2! ways (2×1=22 \times 1 = 2).

  4. Calculate the total number of favorable arrangements (E1E_1).

    E1=(arrangements of blocks)×(arrangements within M)×(arrangements within H)×(arrangements within S)×(arrangements within B)E_1 = (\text{arrangements of blocks}) \times (\text{arrangements within M}) \times (\text{arrangements within H}) \times (\text{arrangements within S}) \times (\text{arrangements within B}) E1=4!×4!×3!×3!×2!E_1 = 4! \times 4! \times 3! \times 3! \times 2! E1=24×24×6×6×2=41,472E_1 = 24 \times 24 \times 6 \times 6 \times 2 = 41,472

  5. Calculate the probability.

    P(E1)=E1N=41,472479,001,600P(E_1) = \frac{E_1}{N} = \frac{41,472}{479,001,600} P(E1)=111,550P(E_1) = \frac{1}{11,550}

    The probability that all books of the same subject land up next to each other is 111,550\boxed{\frac{1}{11,550}}.

Part 23.2

122\frac{1}{22}

Determine the probability that just the History books will be together.

  1. Treat the 3 History books as a single block.

    The other books are 4 Mathematics, 3 Science, and 2 Biology books. These are 4+3+2=94+3+2=9 individual books.

  2. Arrange the block of History books and the 9 other individual books.

    We are arranging a total of 1 (History block)+9 (other books)=101 \text{ (History block)} + 9 \text{ (other books)} = 10 items.
    The number of ways to arrange these 10 items is 10!10!.

    10!=3,628,80010! = 3,628,800

  3. Arrange the books within the History block.

    The 3 History books within their block can be arranged in 3!3! ways.

    3!=3×2×1=63! = 3 \times 2 \times 1 = 6

  4. Calculate the total number of favorable arrangements (E2E_2).

    E2=(arrangements of 10 items)×(arrangements within History block)E_2 = (\text{arrangements of 10 items}) \times (\text{arrangements within History block}) E2=10!×3!E_2 = 10! \times 3! E2=3,628,800×6=21,772,800E_2 = 3,628,800 \times 6 = 21,772,800

  5. Calculate the probability.

    P(E2)=E2N=10!×3!12!P(E_2) = \frac{E_2}{N} = \frac{10! \times 3!}{12!}

    We can simplify this expression:

    P(E2)=10!×3!12×11×10!P(E_2) = \frac{10! \times 3!}{12 \times 11 \times 10!} P(E2)=3!12×11P(E_2) = \frac{3!}{12 \times 11} P(E2)=6132P(E_2) = \frac{6}{132} P(E2)=122P(E_2) = \frac{1}{22}

    The probability that just the History books will be together is 122\boxed{\frac{1}{22}}.

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