Question

The number of ways in which 5 boys and 5 girls can be arranged in a row so that no two girls and no two boys are together is
(a) \[2{{\left( 5! \right)}^{2}}\]
(b) \[{{\left( 5! \right)}^{2}}\]
(c) \[5!6!\]
(d) \[10!\]

Answer 1

Hint: We solve this problem by using the permutations that are arrangements of objects.

Complete answer:
First we arrange the boys in any order in the row.
We have the condition that number of arrangements of \[x\] objects in a row is given as \[x!\]
Then we find the number of gaps between the 5 boys and arrange the girls in that gaps because no two girls should be together means for every two girls there should be at least 1 boy
 We have the condition that number of arrangements of \[x\] objects in \[y\] places is given as \[{}^{y}{{P}_{x}}\] where,
\[{}^{y}{{P}_{x}}=\dfrac{y!}{\left( y-x \right)!}\]
We are given that there are total of 5 boys and 5 girls
We are asked to find the number of arrangements such that no two boys and no two girls are together.
Now, let us arrange the boys first
Let us assume that the number of arrangements of boys as \[B\]
We know that the condition that number of arrangements of \[x\] objects in a row is given as \[x!\]
By using the above condition we get
\[\Rightarrow B=5!\]
Now, let us find the number of places in which girls can be placed.
We are given that no two girls and no two boys are together.
Here, we can see that the only possibility is arranging the girls between the boys.
Here, we can see that we have arranged \[5\] boys so that there are \[6\] places between them.
This is because the first place and the last place can also be occupied by the girls.
Let us assume that the number of arrangements of \[5\] girls as \[G\]
We know that the condition that number of arrangements of \[x\] objects in \[y\] places is given as \[{}^{y}{{P}_{x}}\] where,
\[{}^{y}{{P}_{x}}=\dfrac{y!}{\left( y-x \right)!}\]
By using the above condition we get
\[\Rightarrow G={}^{6}{{P}_{5}}\]
Now, let us assume that the total number of required ways as \[N\]
Here, we know that the total number of ways is the permutations for arrangements of boys and girls.
By using the above condition we get
\[\begin{align}
  & \Rightarrow N=B\times G \\
 & \Rightarrow N=5!\times {}^{6}{{P}_{5}} \\
 & \Rightarrow N=5!\times \dfrac{6!}{\left( 6-5 \right)!} \\
 & \Rightarrow N=5!6! \\
\end{align}\]
Therefore, we can conclude that the total number of ways of such arrangements is \[5!6!\]

So, option (c) is the correct answer.

Note:
Students may make mistakes in taking the number of places that are left for that girl.
Here, we have a total of 6 places for girls after arranging 5 boys.
This is because the first place can also be occupied by a girl so that the number of places will be 6.
But students may do mistake without taking the first place and assume that the total number of places for girls as 5