Question

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

Answer 1

Hint: We can take teachers of each section as 1 set. Then there will be 3 such sets. Each set can be arranged in 2 different ways and 3 such sets can be arranged in \[3!\] different ways. The total number of arrangements is given by the product of all the possible arrangements.

Complete step-by-step answer:
We have 6 teachers.
It is given that 2 are primary teachers. They can stand together in a row in $2!$ ways.
The two middle teachers also can stand in an arrow in $2!$ways.
There are 2 secondary teachers and they can also stand together in a row in $2!$ ways.
Now we have 3 sets of teachers. The 3 sets can be arranged in $3!$ ways.
So the total number of the ways is given by the product of all the possible ways.
 $ \Rightarrow N = 2! \times 2! \times 2! \times 3!$
On expanding the factorial, we get,
 $ \Rightarrow N = 2 \times 2 \times 2 \times 3 \times 2$
On taking the product, we get,
  $ \Rightarrow N = 48$
So the six teachers can be arranged in 48 ways.

Note: We used the concept of permutations to solve the problem. $n!$ represents the number of ways of arranging n objects. The factorial a number n is defined as the product of all the positive integers less than or equal to n.
Mathematically, $n! = n \times \left( {n - 1} \right)\left( {n - 2} \right).... \times 2 \times 1$.
While doing this problem, students do not take a number of ways of arranging the 2 teachers inside the same set. This arrangement must be taken for all the three sets. We must use the product and not the sum to find the total possible arrangement.