Math Problem Statement

7 people visited a marriage function in 3 different vehicles, each accomodating a maximum of 5 persons. In how many ways can they visit such that they use all the 3 vehicles

Solution

The problem is asking for the number of ways to arrange 7 people in 3 vehicles, where each vehicle can accommodate a maximum of 5 people, and all vehicles must be used.

We can solve this using the following steps:

Step 1: Distribute 7 people into 3 vehicles

Let the number of people in each vehicle be x1x_1, x2x_2, and x3x_3, where x1+x2+x3=7x_1 + x_2 + x_3 = 7 and each vehicle has at least 1 person (since all 3 vehicles are used). Additionally, each vehicle can have a maximum of 5 people. So, we need to find the number of integer solutions to:

x1+x2+x3=7x_1 + x_2 + x_3 = 7 where 1x1,x2,x351 \leq x_1, x_2, x_3 \leq 5.

To handle this, let's first transform the variables by letting y1=x11y_1 = x_1 - 1, y2=x21y_2 = x_2 - 1, and y3=x31y_3 = x_3 - 1, so that now:

y1+y2+y3=4y_1 + y_2 + y_3 = 4 where 0y1,y2,y340 \leq y_1, y_2, y_3 \leq 4.

Step 2: Count the number of solutions

This is now a stars and bars problem, where we need to find the number of non-negative integer solutions to y1+y2+y3=4y_1 + y_2 + y_3 = 4, subject to the constraint that each yi4y_i \leq 4.

The number of unrestricted solutions (without the upper limit of 4) is given by the stars and bars formula:

(4+22)=(62)=15.\binom{4+2}{2} = \binom{6}{2} = 15.

Since none of the variables can exceed 4, and this is already within the allowable range, no further adjustments are necessary.

Step 3: Permutations within the groups

Now that we have determined the number of ways to distribute people among the vehicles, we need to account for the number of ways to assign specific people to the vehicles.

After choosing how many people go in each vehicle, we can arrange the 7 people in those vehicles. This is done by permuting the 7 people across the groups. So the number of ways to assign specific people to the groups is:

7!x1!x2!x3!\frac{7!}{x_1!x_2!x_3!}

However, the details of this require further clarification depending on specific numbers of

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Stars and Bars Method
Permutations

Formulas

Stars and Bars formula: \binom{n + k - 1}{k - 1}
Permutations formula: \frac{n!}{x_1!x_2!x_3!}

Theorems

Stars and Bars Theorem
Permutation Theorem

Suitable Grade Level

Grades 11-12