37. Permutations and combinations problems

37. Permutations and combinations problems - step by step solution procedure and detailed explanation for each problem.

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Below are 10 permutations and combinations problems, each with a step-by-step solution procedure and a detailed explanation. The problems are designed to cover a range of concepts, from basic to advanced, to help you understand the applications of permutations and combinations. Each problem includes the problem statement, solution steps, and an explanation of the reasoning and formulas used.
Problem 1: Arranging 3 Books on a Shelf
Problem: You have 3 distinct books. In how many ways can you arrange them on a shelf?
Solution Procedure:
  1. Identify the type of problem: This is a permutation problem because the order of arrangement matters (e.g., ABC is different from CBA).
  2. Determine the formula: For arranging ( n ) distinct objects, the number of permutations is ( n! ) (factorial).
  3. Apply the formula: Here,
    n = 3
    , so calculate
    3! = 3 \times 2 \times 1 = 6
    .
  4. State the answer: There are 6 ways to arrange the books.
Detailed Explanation:
  • Why permutations? Since the books are distinct and the order matters (different arrangements are considered unique), we use permutations.
  • Factorial explanation: The first book has 3 choices, the second has 2 remaining choices, and the third has 1 choice. Thus,
    3 \times 2 \times 1 = 6
    .
  • Verification: List the arrangements: ABC, ACB, BAC, BCA, CAB, CBA (6 total).
Answer: 6 ways.
Problem 2: Choosing 2 Fruits from 5
Problem: You have 5 distinct fruits (apple, banana, orange, mango, peach). In how many ways can you choose 2 fruits?
Solution Procedure:
  1. Identify the type of problem: This is a combination problem because the order of selection does not matter (e.g., choosing apple and banana is the same as banana and apple).
  2. Determine the formula: The number of ways to choose ( r ) items from ( n ) items is given by the combination formula
    C(n, r) = \frac{n!}{r!(n-r)!}
    .
  3. Apply the formula: Here,
    n = 5
    ,
    r = 2
    , so calculate
    C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2! \cdot 3!} = \frac{5 \times 4 \times 3!}{2 \times 1 \cdot 3!} = \frac{5 \times 4}{2 \times 1} = 10
    .
  4. State the answer: There are 10 ways to choose 2 fruits.
Detailed Explanation:
  • Why combinations? The problem asks for choosing fruits without regard to the order of selection, so we use combinations.
  • Formula breakdown: The formula ( C(n, r) ) calculates the number of ways to select ( r ) items from ( n ) without repetition and without order. The factorials cancel out to simplify the calculation.
  • Verification: List the pairs: {Apple, Banana}, {Apple, Orange}, {Apple, Mango}, {Apple, Peach}, {Banana, Orange}, {Banana, Mango}, {Banana, Peach}, {Orange, Mango}, {Orange, Peach}, {Mango, Peach} (10 total).
Answer: 10 ways.
Problem 3: Arranging 4 People in a Line with Restrictions
Problem: In how many ways can 4 distinct people (A, B, C, D) be arranged in a line such that A and B are not adjacent?
Solution Procedure:
  1. Calculate total arrangements: The total number of ways to arrange 4 people is
    4! = 24
    .
  2. Calculate arrangements where A and B are adjacent:
    • Treat A and B as a single "block" or unit. This reduces the problem to arranging 3 units (the AB block, C, D).
    • Number of ways to arrange these 3 units =
      3! = 6
      .
    • Within the AB block, A and B can be arranged in
      2! = 2
      ways (AB or BA).
    • Total arrangements where A and B are adjacent =
      3! \times 2! = 6 \times 2 = 12
      .
  3. Calculate arrangements where A and B are not adjacent: Subtract the adjacent case from the total:
    24 - 12 = 12
    .
  4. State the answer: There are 12 ways to arrange the people such that A and B are not adjacent.
Detailed Explanation:
  • Why permutations? The problem involves arranging people in a line, where order matters.
  • Block method: Treating A and B as a single unit simplifies the calculation of arrangements where they are adjacent. The ( 2! ) accounts for the internal arrangements of A and B within the block.
  • Subtraction: Since we want arrangements where A and B are not adjacent, we subtract the number of arrangements where they are adjacent from the total.
  • Verification: List a few non-adjacent arrangements, e.g., ACBD, ADBC, CADB, etc., and confirm the logic.
Answer: 12 ways.
Problem 4: Forming a Committee with Choices
Problem: A club has 8 members. In how many ways can a committee of 3 members be formed if one specific member (say, Alice) must be included?
Solution Procedure:
  1. Identify the type of problem: This is a combination problem because the order of selection for the committee does not matter.
  2. Account for the restriction: Since Alice must be included, we select her automatically, leaving us to choose 2 more members from the remaining 7 members.
  3. Apply the combination formula: Choose 2 members from 7:
    C(7, 2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2! \cdot 5!} = \frac{7 \times 6}{2 \times 1} = 21
    .
  4. State the answer: There are 21 ways to form the committee.
Detailed Explanation:
  • Why combinations? The committee is a group where the order of selection is irrelevant.
  • Restriction handling: By fixing Alice in the committee, we reduce the problem to choosing the remaining members, which simplifies the calculation.
  • Formula application: The combination formula is used for the remaining selections, ensuring no repetition or order considerations.
Answer: 21 ways.
Problem 5: Arranging Letters of a Word
Problem: How many distinct arrangements can be made with the letters of the word "BOOK"?
Solution Procedure:
  1. Identify the type of problem: This is a permutation problem with repeated items.
  2. Count the letters: The word "BOOK" has 4 letters: B, O, O, K.
  3. Determine the formula: For arranging ( n ) items with repetitions, the number of distinct permutations is
    \frac{n!}{n_1! \cdot n_2! \cdot \ldots}
    , where
    n_i
    is the frequency of each repeated item.
  4. Apply the formula: Total letters
    n = 4
    , with O repeated twice (frequency = 2), B and K once each (frequency = 1). Calculate
    \frac{4!}{2! \cdot 1! \cdot 1!} = \frac{24}{2 \times 1 \times 1} = 12
    .
  5. State the answer: There are 12 distinct arrangements.
Detailed Explanation:
  • Why permutations? We are arranging letters, and different orders produce distinct words.
  • Repetition adjustment: The repeated O’s mean that swapping them does not create a new arrangement, so we divide by ( 2! ) to account for this.
  • Verification: List arrangements: BOOK, BOKO, BOOK, BOKO, OOBK, OOKB, OBOK, OKBO, KBOO, KOBO, KOOB, OKOB (12 total, accounting for duplicates).
Answer: 12 arrangements.
Problem 6: Choosing and Arranging 3 Students
Problem: From a group of 6 students, in how many ways can 3 students be chosen and arranged in a row?
Solution Procedure:
  1. Identify the type of problem: This involves both choosing (combinations) and arranging (permutations).
  2. Break it down:
    • Choose 3 students from 6:
      C(6, 3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3! \cdot 3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20
      .
    • Arrange the 3 chosen students:
      3! = 6
      .
  3. Combine the steps: Multiply the number of ways to choose and arrange:
    20 \times 6 = 120
    .
  4. Alternative approach: Use the permutation formula directly:
    P(6, 3) = \frac{6!}{(6-3)!} = \frac{6!}{3!} = 6 \times 5 \times 4 = 120
    .
  5. State the answer: There are 120 ways.
Detailed Explanation:
  • Why both? Choosing students is order-independent (combinations), but arranging them in a row is order-dependent (permutations).
  • Permutation formula:
    P(n, r) = \frac{n!}{(n-r)!}
    directly computes the number of ways to choose and arrange ( r ) items from ( n ).
  • Equivalence: The two methods (combinations × permutations or direct permutations) are equivalent because
    P(n, r) = C(n, r) \times r!
    .
Answer: 120 ways.
Problem 7: Circular Arrangements
Problem: In how many ways can 5 distinct people be seated around a circular table?
Solution Procedure:
  1. Identify the type of problem: This is a circular permutation problem because arrangements in a circle are considered equivalent under rotation.
  2. Determine the formula: For ( n ) distinct objects in a circular arrangement, the number of ways is
    (n-1)!
    .
  3. Apply the formula: Here,
    n = 5
    , so calculate
    (5-1)! = 4! = 24
    .
  4. State the answer: There are 24 ways to arrange the people.
Detailed Explanation:
  • Why circular? In a circular arrangement, rotating the table does not create a new arrangement (e.g., ABCDE is the same as BCDEA).
  • Formula derivation: Fixing one person in a position (to account for rotations) leaves
    (n-1)
    positions to arrange the remaining people, hence
    (n-1)!
    .
  • Verification: For smaller cases, e.g.,
    n=3
    ,
    (3-1)! = 2
    , which matches manual listing (e.g., ABC, ACB).
Answer: 24 ways.
Problem 8: Distributing Identical Objects
Problem: In how many ways can 5 identical candies be distributed among 3 distinct children?
Solution Procedure:
  1. Identify the type of problem: This is a combinations problem involving distributing identical items into distinct groups (stars and bars theorem).
  2. Determine the formula: The number of ways to distribute ( n ) identical items into ( k ) distinct groups (where each group can receive zero or more items) is
    C(n+k-1, k-1)
    .
  3. Apply the formula: Here,
    n = 5
    (candies),
    k = 3
    (children). Calculate
    C(5+3-1, 3-1) = C(7, 2) = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21
    .
  4. State the answer: There are 21 ways to distribute the candies.
Detailed Explanation:
  • Stars and bars: Represent the 5 candies as stars (*) and use 2 bars (|) to separate them into 3 groups. The number of ways to arrange 5 stars and 2 bars is
    C(5+2, 2) = C(7, 2)
    .
  • Example: One arrangement might be ||*** (2 candies to the first child, 0 to the second, 3 to the third).
  • Verification: List distributions like (5,0,0), (4,1,0), (3,2,0), etc., and count 21 total.
Answer: 21 ways.
Problem 9: Forming a Team with Restrictions
Problem: A team of 4 people must be formed from 5 men and 3 women. How many ways can this be done if the team must include at least 1 woman?
Solution Procedure:
  1. Identify the type of problem: This is a combination problem with a restriction.
  2. Calculate total ways to form the team: Choose 4 people from
    5 + 3 = 8
    :
    C(8, 4) = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70
    .
  3. Calculate ways with no women: Choose 4 people from 5 men:
    C(5, 4) = \frac{5!}{4!1!} = 5
    .
  4. Calculate ways with at least 1 woman: Subtract the no-women case from the total:
    70 - 5 = 65
    .
  5. State the answer: There are 65 ways to form the team.
Detailed Explanation:
  • Why combinations? The team is a group, so order does not matter.
  • Subtraction method: The total number of teams includes those with and without women. Subtracting teams with only men gives teams with at least one woman.
  • Alternative: Sum cases (1 woman + 3 men, 2 women + 2 men, 3 women + 1 man):
    • 1 woman:
      C(3, 1) \times C(5, 3) = 3 \times 10 = 30
      .
    • 2 women:
      C(3, 2) \times C(5, 2) = 3 \times 10 = 30
      .
    • 3 women:
      C(3, 3) \times C(5, 1) = 1 \times 5 = 5
      .
    • Total:
      30 + 30 + 5 = 65
      .
Answer: 65 ways.
Problem 10: Arranging People with Identical Twins
Problem: In how many ways can 6 people, including a pair of identical twins, be arranged in a line?
Solution Procedure:
  1. Identify the type of problem: This is a permutation problem with identical items.
  2. Count the people: There are 6 people, but the twins are identical.
  3. Apply the formula: For ( n ) items with repetitions, use
    \frac{n!}{n_1! \cdot n_2! \cdot \ldots}
    . Here,
    n = 6
    , with the twins repeated (frequency = 2), and the other 4 people distinct (frequency = 1 each). Calculate
    \frac{6!}{2! \cdot 1! \cdot 1! \cdot 1! \cdot 1!} = \frac{720}{2} = 360
    .
  4. State the answer: There are 360 ways to arrange the people.
Detailed Explanation:
  • Why permutations? The problem involves arranging people in a line, where order matters.
  • Identical twins: The twins being identical means swapping them does not create a new arrangement, so we divide by ( 2! ).
  • Verification: Consider a smaller case with 3 people (2 identical, 1 distinct):
    \frac{3!}{2!} = 3
    , which matches arrangements like T1T2A, T1AT2, AT1T2.
Answer: 360 ways.
Summary of Key Concepts
  • Permutations: Used when order matters (e.g., arrangements). Formula:
    P(n, r) = \frac{n!}{(n-r)!}
    or ( n! ) for full arrangements.
  • Combinations: Used when order does not matter (e.g., selections). Formula:
    C(n, r) = \frac{n!}{r!(n-r)!}
    .
  • Repetitions: Divide by factorials of frequencies for identical items.
  • Restrictions: Use subtraction (total minus restricted cases) or case-by-case analysis.
  • Circular arrangements: Use
    (n-1)!
    to account for rotational symmetry.
  • Stars and bars: Used for distributing identical items.
These problems cover a broad spectrum of permutations and combinations scenarios, with detailed explanations to clarify the reasoning and calculations. 

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