Question

In how many ways may n girls and n boys be seated in a row of 2n chairs, If no two boys or girls must sit together.

Answer 1

Hint: For solving this question, first we will see the important concepts like the fundamental principle of multiplication from permutations and combinations. After that, we will make n girls be seated alternatively in the row and arrange them in $n!$ ways. Then, we will make boys to be seated in the empty spaces between girls and use the formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ to get the number of ways in which boys can be seated. Then, we will multiply both results to get the total number of ways possible.

Complete step-by-step answer:
Given: - We have to find the number of ways in which n girls and n boys be seated in a row so that no two boys are together.
Now, before we proceed, we should know the following important concept and formulas:
1. Fundamental Principle of Multiplication: If there are two jobs such that one of them can be completed in p ways, and when it has been completed in any of these p ways, the second job can be completed in q ways. Then, two jobs in succession can be completed in $p \times q$ ways.
2. Number of ways to select r objects from the n distinct objects. The formula for the number of different possible ways is $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
3. Number of linear arrangements of r distinct objects will be equal to $r!$.
We can arrange the n girls in a total $n!$.
Now, if we make boys be seated in the empty n spaces then, no two boys will be sitting together. And as we know that, we can select r objects from n distinct objects in $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
We can arrange the n boys in a total $n!$.
Now, from the fundamental principle of multiplication, we can say that the number of ways in which n girls and n boys be seated in a row so that two sex must alternate will be equal to,
$n! \times n!$
Since the order of seating can be either boy seat on the first chair or girl. So,
$\therefore $Total ways $ = 2\left( {n!} \right)\left( {n!} \right)$.

Hence, the total number of ways is $2\left( {n!} \right)\left( {n!} \right)$.

Note: And in such questions of permutations and combinations. Apply the basic concepts and try to visualize the given event through figures. Order of events is important in permutations. Order of events is not important in combinations.