Question

There are three girls in a class of 10 students. The number of different ways in which they can be seated in a row such that no two of the three girls are together is
A) \[7!{\text{ }} \times {}^6P_3\]
B) \[7!{\text{ }} \times {}^8P_3\]
C) \[7!{\text{ }} \times 3!\]
D) \[\dfrac{{10!}}{{3!7!}}\]

Answer 1

Hint: Here first we will find the number of boys and then find the number of ways in which the boys can be arranged in a row and then we will find the number of ways in which the number of girls can be arranged in remaining number of places in a row such that no two girls are together.

Complete step by step answer:
Since we are given that there are 3 girls in a class of 10 students hence,
The number of boys in the class \[ = 10 - 3\]
                                                           \[ = 7\]
Now, we have to find the number of ways in which these boys can be arranged in a row leaving a space for a girl between two boys as two girls can’t be seated together.

B

B

B

B

B

B

B

Seven boys can be arranged in a row in \[{}^7P_7\] ways
Since we know that,
\[{}^nP_r = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Therefore,
The number of ways in which seven boys can be arranged in a row is given by:
\[
   = \dfrac{{7!}}{{\left( {7 - 7} \right)!}} \\
   = \dfrac{{7!}}{{0!}} \\
   = 7! \\
 \]
The value of \[0!\] is 1
Therefore the boys can be arranged in 7! Ways
Now we need to find the number of ways in which the girls can be be arranged in the remaining empty spaces hence,
Total number of empty spaces = 8
Number of girls = 3
Therefore, the number of ways in which 3 girls can be arranged at 8 places is given by:
\[{}^8P_3\] ways

Now, the total number of ways in which both boys and girls can be arranged in a row \[ = 7!{\text{ }} \times {}^8P_3\]. Hence option B is the correct option.

Note:
The formula for permutation(arrangement) of r numbers out of n numbers is given by:
\[{}^nP_r = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Also, the value of \[0!\] is 1.
Since the girls and boys need to be arranged , therefore permutation should be used and not combined.