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?
- A6
- B12Correct
- C18
- D24
Explanation
To solve this, we can treat the two children A and B as a single unit because they must always be together.
First, consider the total number of entities to arrange. Since A and B are grouped together, we have three entities: the AB unit, child C, and child D. The number of ways to arrange these three entities in a line is 3 factorial, which equals 3 times 2 times 1, resulting in 6 ways.
Second, within the AB unit itself, A and B can switch places. They can stand as AB or BA. This gives us 2 internal arrangements.
Finally, to find the total number of ways, we multiply the arrangements of the entities by the internal arrangements of the unit. This is 6 multiplied by 2, which equals 12.
Therefore, there are 12 different ways the children can be arranged. Option B is the correct answer.
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