In how many ways can four children be made to stand in a line such that two of them , A and B are always together ?

Options

A. A)6
B. B)12
C. C)18
D. D)24

Answer

To solve the problem of how many ways four children can stand in a line with the condition that two of them, A and B, are always together, we can follow these steps:

### Step 1: Treat A and B as a single unit
Since A and B must always be together, we can consider them as one unit or block. This means instead of having four individual children (A, B, C, D), we now have three units to arrange: (AB), C, and D.

**Hint:** When two items must stay together, group them as a single unit to simplify the arrangement.

### Step 2: Count the total units
Now we have three units: (AB), C, and D.

**Hint:** Count the total number of units after grouping the items that must stay together.

### Step 3: Calculate the arrangements of the units
The number of ways to arrange these three units is given by the factorial of the number of units. Thus, we calculate:
\[ 3! = 3 \times 2 \times 1 = 6 \]

**Hint:** Use factorial to find the number of ways to arrange distinct units.

### Step 4: Consider the arrangements within the unit (AB)
Since A and B can switch places within their block, we need to account for the arrangements of A and B. There are 2 ways to arrange A and B (AB or BA).

**Hint:** For any unit that can be internally arranged, calculate the arrangements separately.

### Step 5: Multiply the arrangements
Now, to find the total arrangements, we multiply the arrangements of the units by the arrangements within the block:
\[ \text{Total arrangements} = 3! \times 2! = 6 \times 2 = 12 \]

**Hint:** Combine the arrangements of the grouped units with the arrangements within the group to get the final count.

### Conclusion
Thus, the total number of ways the four children can stand in line such that A and B are always together is **12**.

**Final Answer:** 12