Question

The number of ways in which 5 girls and 3 boys can be seated in a row so that no two boys are together is?
(a) 14040
(b) 14440
(c) 14000
(d) 14400

Answer 1

We start solving the problem by finding the way that the boys and girls can be arranged according to the condition given in the problem. We then find the total no. of ways of arranging the five girls. We then find the total number of ways of arranging the boys in such a way that no two boys are together. We then multiply both of them to get the number of ways such that the boys and girls can be seated as mentioned in the problem.

Complete step-by-step solution:
According to the problem, we need to find the number of ways in which 5 girls and 3 boys can be seated in a row such that no two boys sit together.
From the problem, we can understand that the girls can be seated in any way they need but the boys must not be together. This can only happen if one girl is present in between two boys. We first make the girls seated in such a way that there is one gap on both sides of them and we arrange the 3 boys in those places.
Let us first find the total number of ways that 5 girls can be seated as shown below.

| G| | G| | G| | G| | G|
---|---|---|---|---|---|---|---|---|---|---

We can see that the seating of girls resembles the arrangement of ‘n’ objects in ‘n’ places which is $n!$ ways.
So, the 5 girls can be seated in $5!$ ways.
We know that $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 3\times 2\times 1$.
So, we have $5!=5\times 4\times 3\times 2\times 1=120$. This tells us that 5 girls can be seated in 120 ways -(1).
Now, we need to arrange 3 boys in those 6 vacant places. We can see that this is similar to arranging ‘r’ objects in ‘n’ places $\left( r < n \right)$ which can be done in ${}^{n}{{P}_{r}}$ ways.
So, the boys can be arranged in ${}^{6}{{P}_{3}}$ ways.
We know that ${}^{n}{{P}_{r}}=n\times \left( n-1 \right)\times \left( n-2 \right)\times .....\times \left( n-r+1 \right)$.
So, we have ${}^{6}{{P}_{3}}=6\times 5\times 4=120$ -(2).
Since we need to find the total number of ways that girls and boys sit in this way, we multiply the results obtained in (1) and (2).
So, the total number of ways is $120\times 120=14400$.
We have found that the number of ways in which 5 girls and 3 boys can be seated in a row so that no two boys are together is 14400.
The correct option for the given problem is (d).

Note: Whenever we have these types of problems, we first try to find a feasible seating position for both boys and girls as this will give us the idea required for the answer. We can also find the total ways of arranging the boy by first choosing the places for all three boys and then arranging them in those places using the combinations. We can also find the probability for this case by dividing the obtained answer by the total number of ways of arranging 5 girls and 3 boys. Similarly, we can expect problems to find the total number of ways that all the boys and girls sit together.