Question

In how many ways can 5 girls and 3 boys stand in a row so that no two boys are together?
A. 14560
B. 14300
C. 14400
D. 16400

Answer 1

Hint: In general permutation can be defined as each of several possible ways in which a set or number of things can be ordered or arranged. The main difference between permutations and combination is that when the order of arranging the things doesn’t matter it is a combination and when the order does matter it is a permutation.
The number of permutations of n objects taken r at a time determined by the formula
\[{}^{n}{{P}_{r}}=P(n,r)=\dfrac{n!}{(n-r)!}\]
n = total number of objects
r = number of objects selected.

Complete step by step answer:
From our question we have 5 girls and 3 boys that need to be arranged with the condition that no two boys are together.
So it can be said there should be a at least one girl between two boys
Let G represent the girls and B represent the boys
Our arrangement can be represented as
_ G _ G _ G _ G _ G _
so that there will always be at least one girl between two boys.
Here we should calculate two types of permutations
(i) Arranging the 5 girls in the 5 positions.
(ii) Arranging 3 boys from the above 6 places(the blanks)
So first let us arrange the 5 girls in 5 positions
From our formula we have
n = 5, r = 5
\[\begin{align}
  & {}^{n}{{P}_{r}}=P(5,5)=\dfrac{5!}{(5-5)!}=\dfrac{5!}{0!}=\dfrac{5\times 4\times 3\times 2\times 1}{1} \\
 & {}^{n}{{P}_{r}}={}^{5}{{P}_{5}}=120 \\
\end{align}\]
 Now let us arrange the 3 boys in 6 places
n = 6, r = 3
\[\begin{align}
  & {}^{n}{{P}_{r}}=P(6,3)=\dfrac{6!}{(6-3)!}=\dfrac{6!}{3!}=\dfrac{6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1}=\dfrac{720}{6} \\
 & {}^{n}{{P}_{r}}={}^{6}{{P}_{3}}=120 \\
\end{align}\]
Total number of permutations = 120 $\times$ 120 = 14400

So, the correct answer is “Option C”.

Note: Students solving this problem shouldn’t forget to add up both the permutations at the end. Most frequently the arrangement of girls wouldn’t be done by the student. This problem can also be solved by subtracting permutations of two boys and three boys together from the total number of permutations of 5 girls and 3 boys.