The number of ways in which 5 male and 2 female members of committee can be seated around a table, so that two female members are not seated by together is
A. 480
B. 600
C. 720
D. 840

Hint: We will start by fixing one male at a particular position and find the remaining number of ways of 4 members then find the number of ways where two females cannot sit together out of 5 available seats. We will use the permutation formula \[{}^n{p_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] to expand the value.

Complete step by step solution:
It is given that there are 5 male and 2 female members seated around a table such that two female members are not seated together.
Initially we have to fix up a male at a particular position.
Since the number of positions is 5, out of which one seat is fixed for one male member already, hence the remaining 4 males can be seated in 4! ways.
As no two females are seated together and if we take any two males, there is one place between them. Therefore, for 5 males there are 5 empty seats available between two consecutive males.
So, the possible arrangement will be ${}^5{P_2}$
Therefore, the total number of ways can be found by multiplying the remaining number of ways with possible arrangements, that is, \[4!{\text{ }} \times {}^5{P_2}\]
Thus, we have solved the value of ${}^5{P_2}$ using the formula,
We get,
$
   \Rightarrow {}^5{P_2} = \dfrac{{5!}}{{3!}} \\
   \Rightarrow {}^5{P_2} = \dfrac{{5 \times 4 \times 3!}}{{3!}} \\
   \Rightarrow {}^5{P_2} = 20 \\
$
Now, we will find the product to determine the number of ways such that no two females are seated together.
$
   = \,4 \times 3 \times 2 \times 1 \times 20 \\
   = 480\,{\text{ways}} \\
$
Hence, the correct option is A.

Note: Apply permutation where total number of possibilities are given out of which particular possibilities are asked. Apply the permutation formula that is \[{}^n{p_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] where \[n\] represents the total number of members. As one man has selected one position then the number of choices left for 2 people is 4 and in a similar way for 3 people the number of choices left is 3 and so on.