QUESTION

# Three boys of class X, four boys of class XI, and five boys of class XII sit in a row. The total number of ways in which these boys can sit so that all the boys of the same class sit together is equal to:(a) ${{\left( 3! \right)}^{2}}\left( 4! \right)\left( 5! \right)$(b) $\left( 3! \right){{\left( 4! \right)}^{2}}\left( 5! \right)$(c) $\left( 3! \right)\left( 4! \right)\left( 5! \right)$(d) $\left( 3! \right)\left( 4! \right){{\left( 5! \right)}^{2}}$

Hint: For solving this problem we will break down the given condition into simple cases and then use the formula of the number of different ways in which we can arrange a certain number of different objects.

Given:
There are a total of 12 boys in which, 3 boys are of class X, 4 boys are of class XI and 5 boys are of class XII. And these 12 boys are sitting in a row such that all the boys of the same class sit together.

We have to find the number of different ways in which 12 boys can sit as per the given condition. Now, before we proceed to solve the question we should know in how many different ways we can arrange $N$ different objects in a linear combination.

If we have $N$ different objects then, we can arrange them in a linear arrangement in $N!$ different ways.

As it is given in the question that all boys of the same class sit together. Then, we can just club the different boys of the same class into one group. In other words, if we have to arrange 12 boys then, first we will club 3 boys of class X into 1 group of class X and similarly one more group for class XI boys and one more group for class XII boys.

We have, 3 groups as follow:
First group of 3 boys of class X.
Second group of 4 boys of class XI.
Third group of 5 boys of class XII.

Now, we have to arrange these 3 groups in a linear arrangement then, we can arrange them in $3!$ different ways.

Then, we can also arrange boys within their group such that they can sit together. For example:

In the first group of 3 boys of class X, we can arrange these boys within their group in $3!$ different ways. And then in the second group of 4 boys of class XI, we can arrange these boys within their groups in $4!$ different ways. Similarly in the third group of 5 boys of class XII, we can arrange these boys within their group in $5!$ different ways.

Now, to find the total number of different ways in which 12 boys can sit in a row such that boys of the same class sit together can be calculated by multiplying all the different numbers of possible arrangements that we have calculated above. Then,

Total number of different arrangements $=\left( 3! \right)\times \left( 3! \right)\times \left( 4! \right)\times \left( 5! \right)={{\left( 3! \right)}^{2}}\left( 4! \right)\left( 5! \right)$ .

Hence, (a) is the correct option.

Note: Here, the student must take care while making different cases that are possible and not directly apply the formula for the total number of different arrangements of a certain number of different objects in a linear arrangement.