Math Problem Statement

two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated?

Solution

To solve this problem, we need to determine the number of ways to seat three families in a row such that the members of each family are seated together.

Step-by-Step Solution

  1. Treat each family as a single unit or "block":

    Since the members of each family must be seated together, we can treat each family as a single "block" or "unit." Let's denote the two families with three members each as F1F_1 and F2F_2, and the family with four members as F3F_3.

    Thus, we have three "blocks" to arrange: F1F_1, F2F_2, and F3F_3.

  2. Arrange the "blocks" (families) in a row:

    There are 3 families (or "blocks"), so they can be arranged in a row in: 3!=6 ways.3! = 6 \text{ ways.}

  3. Arrange the members within each family:

    Now, for each family block, we need to arrange the members within that block:

    • For F1F_1 (3 members), there are: 3!=6 ways to arrange.3! = 6 \text{ ways to arrange.}

    • For F2F_2 (3 members), there are: 3!=6 ways to arrange.3! = 6 \text{ ways to arrange.}

    • For F3F_3 (4 members), there are: 4!=24 ways to arrange.4! = 24 \text{ ways to arrange.}

  4. Combine all the arrangements:

    The total number of ways to seat the families so that members of the same family are not separated is obtained by multiplying the number of arrangements of the families (blocks) with the arrangements of members within each family:

    3!×3!×3!×4!=6×6×6×24=5184.3! \times 3! \times 3! \times 4! = 6 \times 6 \times 6 \times 24 = 5184.

Final Answer

Thus, the total number of ways to seat the families so that the members of each family are not separated is 5184.


Would you like more details or have any questions? Here are some related questions you might find interesting:

  1. What if there were two families with four members each?
  2. How would the answer change if the families could be seated in a circular arrangement?
  3. What if the restriction was that only two families should not be separated?
  4. How many ways are there if the members of a specific family must always sit in the same order?
  5. How does this change if there is an additional constraint, such as family members sitting alternately?

Tip: When dealing with grouping or seating problems, try to break them down into "blocks" to simplify the arrangement process.

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Math Problem Analysis

Mathematical Concepts

Permutations
Combinatorics
Factorials

Formulas

n! (Factorial): The number of ways to arrange n distinct items in order.
Product of permutations for each family: 3! for two families with three members, 4! for the family with four members.

Theorems

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Suitable Grade Level

Grades 9-12