Math Problem Statement

7 people visited a marriage function in 3 different vehicles each accomodating max of 5 ppl how many ways can they visit such that they use all 3 vehicles

Solution

We are asked to find how many ways 7 people can visit a marriage function using 3 vehicles, each with a capacity of 5 people, such that all 3 vehicles are used.

Step 1: Partitioning the 7 people into 3 groups

We need to split 7 people into 3 groups such that none of the groups is empty (since all 3 vehicles must be used). The groups can have the following distribution:

  • One group with 3 people, one group with 2 people, and one group with 2 people.

Step 2: Counting the number of ways to partition the people

We calculate the number of ways to assign people to these groups.

  1. Choose 3 people for the first group: There are (73)\binom{7}{3} ways to choose 3 people.

    (73)=7×6×53×2×1=35\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

  2. Choose 2 people for the second group: After choosing 3 people for the first group, we are left with 4 people. There are (42)\binom{4}{2} ways to choose 2 people for the second group.

    (42)=4×32×1=6\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6

  3. The remaining 2 people automatically go to the third group: There's only 1 way to assign these last 2 people.

Step 3: Considering the arrangement of vehicles

Since the vehicles are distinct, the arrangement of the groups into the vehicles matters. There are 3! (factorial) ways to assign the 3 groups to the 3 vehicles:

3!=63! = 6

Step 4: Final calculation

Now, multiply all the factors together:

Total ways=(73)×(42)×3!=35×6×6=1260\text{Total ways} = \binom{7}{3} \times \binom{4}{2} \times 3! = 35 \times 6 \times 6 = 1260

Thus, the total number of ways the 7 people can visit using all 3 vehicles is 1260.


Would you like more details on any of the steps?

Related Questions:

  1. How do you calculate the binomial coefficient (nr)\binom{n}{r}?
  2. What is the principle behind arranging groups into distinct sets?
  3. How would the answer change if one vehicle could hold only 3 people?
  4. Can you solve similar problems where some vehicles have larger capacities?
  5. How does Stirling’s formula approximate these types of combinatorial problems?

Tip:

When solving combinatorics problems, breaking the task into smaller, manageable steps—like grouping and arranging—is key to avoiding mistakes!

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Permutations
Combinations
Factorials

Formulas

Binomial Coefficient: \(\binom{n}{r} = \frac{n!}{r!(n - r)!}\)
Factorial: \(n! = n \times (n-1) \times ... \times 1\)

Theorems

Basic Permutation and Combination Theorems

Suitable Grade Level

Grades 9-12