There are six teachers. Out of them two are primary teachers and two are secondary teachers. They are to stand in a row, so as the primary teachers, middle teachers and secondary teachers are always in a set. The number of ways in which they can do so, is

### Concept & Permutations of Multiple Groups When multiple distinct groups must remain together, treat each entire group as a single super-unit. Calculate the permutations of the super-units, and multiply by the internal permutations within every individual super-unit. ### Step-by-Step Solution * **Identify the Sets of Teachers:** * Total teachers = 6. * Primary teachers = 2. * Secondary teachers = 2. * Middle teachers = Total - Primary - Secondary = $6 - 2 - 2 = 2$. * **Form the Groups:** * Group 1: Primary teachers (2 people) * Group 2: Middle teachers (2 people) * Group 3: Secondary teachers (2 people) * **Arrange the Groups:** * There are 3 distinct groups. They can be arranged in a row in $3!$ ways. * $3! = 6$ ways. * **Arrange Teachers Internally Within Groups:** * Primary teachers can swap places: $2! = 2$ ways. * Middle teachers can swap places: $2! = 2$ ways. * Secondary teachers can swap places: $2! = 2$ ways. * **Calculate Total Arrangements:** * Total ways = $3! \times 2! \times 2! \times 2!$ * Total ways = $6 \times 2 \times 2 \times 2 = 48$. ### Exam Strategy & Shortcut Recognize the macro/micro structure instantly. 3 blocks $\rightarrow$ $3! = 6$. Each block has 2 items $\rightarrow$ $2^3 = 8$. Total $= 6 \times 8 = 48$. ### Common Pitfall A common mistake is forgetting to deduce the number of middle teachers. While it isn't explicitly stated as "two", basic arithmetic ($6 - 2 - 2$) is required to realize there are 2 middle teachers making up the final set. ### Final Answer Therefore, the correct answer is **48**.