A family consisting of 2 parents and 5 children is to pose for a picture with 4 family members in the front and 3 in the back.How many arrangements are possible if the parents must be next to each other?

We are asked to find how many possible arrangements there are for a family of 2 parents and 5 children to pose for a picture, where:

- 4 family members are in the front row, and 3 family members are in the back row.
- The parents must stand next to each other.

### Step 1: Consider the parents as a block.
Since the parents must always stand next to each other, we can treat them as a single unit, or "block." This reduces the problem from arranging 7 people to arranging 6 units (5 children + 1 parent block).

### Step 2: Count the possible row arrangements.
#### Front Row:
- We are choosing 4 family members to be in the front row. The parent block must occupy 2 adjacent positions in this row.
- There are 3 ways to place this parent block in the front row:
  - Positions 1 and 2,
  - Positions 2 and 3,
  - Positions 3 and 4.
  
- After placing the parent block, we need to fill the remaining 2 spots in the front row with children. There are \( \binom{5}{2} = 10 \) ways to choose 2 children from the 5 to be placed in the front row.

#### Back Row:
- Once the front row is determined, the remaining 3 family members (who are all children) automatically go to the back row. There is only 1 way to assign them to the back row after the front row is set.

### Step 3: Arrange people within the rows.
#### Front Row:
- In each of the 3 positions for the parent block, the parents themselves can be arranged in \( 2! = 2 \) ways.
- The 2 chosen children for the front row can also be arranged in \( 2! = 2 \) ways.

#### Back Row:
- The 3 children in the back row can be arranged in \( 3! = 6 \) ways.

### Step 4: Calculate the total number of arrangements.
We now multiply all the possibilities:

\[
3 \times \binom{5}{2} \times 2! \times 2! \times 3! = 3 \times 10 \times 2 \times 2 \times 6 = 1440
\]

### Final Answer:
There are 1,440 possible arrangements of the family where the parents must stand next to each other.

---

Would you like further details or have any questions about this?

Here are 5 related questions:
1. How many arrangements are possible if the parents do not need to be next to each other?
2. What if the front row must contain exactly 1 parent and 3 children?
3. How many arrangements are possible if no family members are restricted in their positions?
4. What if one child insists on standing in the front row?
5. How many ways are there to choose the positions if all family members are randomly placed?

**Tip:** Treating groups that must stay together as a single unit is a useful strategy in many arrangement problems!

A family consisting of 2 parents and 5 children is to pose for a picture with 4 family members in the front and 3 in the back. How many arrangements are possible if the parents must be next to each other?

Learn how to calculate the number of possible arrangements for a family of 2 parents and 5 children to pose for a picture, with 4 family members in the front and 3 in the back. This problem involves combinatorial methods, including permutations and combinations, and considers the case where the parents must always stand together. Suitable for Grades 10-12 students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation