In how many ways can 5 children be arranged in a line such that two particular children of them are always together

Answer

To solve the problem of arranging 5 children in a line such that two particular children are always together, we can follow these steps:

### Step 1: Treat the two particular children as a single unit
Since we want the two particular children to always be together, we can group them together as one unit. Let's denote these two children as A and B. By treating A and B as a single unit, we can represent the arrangement as follows:
- (AB), C, D, E

Now, instead of arranging 5 individual children, we are arranging 4 units: (AB), C, D, and E.

### Step 2: Calculate the arrangements of the 4 units
The number of ways to arrange these 4 units is given by the factorial of the number of units. Therefore, the number of arrangements of the 4 units is:
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]

### Step 3: Consider the arrangements within the unit (AB)
Within the unit (AB), the two children A and B can be arranged in 2 different ways: either A can be first and B second, or B can be first and A second. Therefore, we have:
\[ 2! = 2 \]

### Step 4: Calculate the total arrangements
To find the total number of arrangements where A and B are always together, we multiply the number of arrangements of the 4 units by the arrangements of A and B within their unit:
\[ \text{Total arrangements} = 4! \times 2! = 24 \times 2 = 48 \]

Thus, the total number of ways to arrange the 5 children such that the two particular children are always together is **48**.